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The Chemical Industry Flashcards

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1
Q

What is the reaction rate ?

A

The change in the amount of reactants or products per unit time.

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2
Q

What are the units for change in concentration per unit time ?

A

mol dm^-3 s^-1

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3
Q

How do you measure the amount of a product or reactant over the curse of the reaction ?

A

By continuous monitering

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4
Q

How do you follow the rate of reaction ?

A
  • pH measurement
  • Gas volume
  • Loss of mass
  • Colour change
  • Titration
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5
Q

How do you use the pH to measure the rate of a reaction ?

A

If the reactants or products is a acid or base, you can follow the pH of the mixture. You can do this by using a pH probe or a pH meter connected to a data logger.

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6
Q

How do you convert the pH data to concentration ?

A

Use the equation > [H+] = 10^-pH

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7
Q

How do you use the Gas volume to measure the rate of reaction ?

A

If a gas is given off, you would collect it in a gas syringe and record how much you’ve got at a regular time interval.

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8
Q

How do you use the loss of mass to measure the rate of reaction ?

A

If a gas is given off, the system will lose mass. You can measure this using a balance.

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9
Q

How do you use colour change to measure the rate of reaction ?

A

If the reactants or products are coloured then you can rack the colour change using a calorimeter. A calorimeter measures the absorbance of the solution. The more concentrated the colour of the solution, the higher the absorbance is.

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10
Q

How do you use a titration to measure the rate of reaction ?

A

You can measure the concentration of a reactant or product in a solution by taking small samples of the reaction mixture at regular intervals and titrating them. Once you have taken the sample, you need to slow down the reaction happening in it. You can do this by diluting the sample with deionised water, or add another chemical that will stop the reaction from happening.

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11
Q

How can you work out the reaction rate ?

A

Concentration-Time graph

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12
Q

How do you plot the concentration-time graph ?

A
  • Repeatedly measuring the concentration of one of the reactants or products over the course of a reaction.
  • The rate at any point of the reaction is given by the gradient at the point on the concentration-time graph.
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13
Q

What do you have to do if the concentration-time graph is a curve ?

A

Draw a tangent to the curve and find the gradient to that

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14
Q

What is the rate of reaction ?

A

The change in the amount of reactants or products per unit time.

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15
Q

Describe how you could use a calorimeter to follow the rate of reaction with one coloured reactant ?

A

The more concentrated the colour of the solution is, the higher the absorbance.

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16
Q

How can a concentration-time graph be used to work out the rate of reaction at a particular point ?

A

It is given using the gradient at that point on the concentration-time graph.

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17
Q

The rate of the acid-catalysed reaction between bromine, Br2, and methanoic acid, HCOOH, was investigated.

Br2 + HCOOH –> 2H+ + 2Br- + CO2

a) Describe one method that could be used to follow the reaction rate ?

b) If the concentration of Br2, was recorded at set intervals over the course of the reaction, outline how the rate of reaction, at any particular time, could be determined.

A

a)
- Measure the volume of CO2 produced using a gas syringe.
- Measure how the colour of the solution changes over time using a calorimeter.
- Measure the decrease in mass using a mass balance.

b) Plot a graph of concentration of Br2 against time. Draw a tangent to the curve at a particular time. Calculate the gradient of the tangent.

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18
Q

What does the initial rates method tell
you ?

A

The rate at the beginning of a reaction

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19
Q

What do you do in the initial rates
reaction ?

A

You time how long it takes for a set amount of product to form at the beginning of the reaction and then use this data in to calculate the initial rate of the reaction.
You repeat the experiment several times, each time changing the initial concentration of one of the reactants.

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20
Q

When you plan to use the initial rates method on an experiment, you have to make the assumptions that … ?

A
  • The concentration of the other reactants isn’t changing significantly > you usually do this by having all the other reactants present in excess.
  • The temperature stays constant
  • The reaction has not proceeded too far when you take your measurement.
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21
Q

What is the initial rate equation ?

A

Initial rate = (amount of reactant used or product formed) / (time)

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22
Q

How can you measure the initial rate of a reaction from a concentration-time
graph ?

A

Draw a tangent to the curve at t = 0 and measure the gradient.

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23
Q

What is an example of an initial rate experiment ?

A

Clock reactions

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24
Q

What is a clock reaction ?

A

You measure how the time taken for a set amount of product to form changes as you vary the concentration of one of the reactants.

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25
Q

What is an observable end point of a clock reaction ?

A

A colour change, which tells you when the desired amount of product has formed.

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26
Q

If the clock reactions finishes quickly, what does this suggest about the initial rate of the reaction ?

A

There is a fast initial rate of reaction.

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27
Q

Describe the Iodine clock reaction.
H2O + 2I- + 2H+ –> 2H2O + I2

A
  • A small amount of sodium thiosulfate solution and starch are added to an excess of hydrogen peroxide and iodide ions in acid solution.
  • The sodium thiosulfate that is added to the reaction mixture instantaneously with any iodine that forms.
    2S2O3^2- + I2 –> 2I- + S4O6^2-
  • To begin with, all the iodine that forms in the first reaction is used up straight away in the second reaction. But once all the solution thiosulfate is used up, any more iodine that forms will stay in solution, so the starch indicator will suddenly turn blue-black. This is the end point of the reaction.
  • Varying the concentration of the iodide ions or hydrogen peroxide will give different times for the colour change.
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28
Q

State two assumptions you need to make when carrying out an initial rates
reaction ?

A
  • The temperature remains constant
  • The concentration of the other reactants isn’t changing significantly.
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29
Q

What happens at the end point of a clock reaction ?

A

Usually a colour change

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30
Q

What does the rate equation link
together ?

A

The reaction rate and the reactant concentrations

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31
Q

What is the rate equation ?

A

Rate = k[A]^m [B]^n

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32
Q

In the rate equation, k[A]^m [B]^n, what does n and m mean ?

A

The orders of reaction with respect to reactant A and reactant B.
(m tells you how the concentration of reactant A affects the rate and n tells you the same for reactant B).

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33
Q

What does k mean in the rate equation ?
Rate = k[A]^m [B]^n

A

The rate constant.

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34
Q

If the rate constant is bigger what does this tell you about the reaction ?

A

The reaction is faster.

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35
Q

When is the rate constant the same ?

A

For a certain reaction at a particular temperature.

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36
Q

The chemical equation below shows the acid-catalysed reaction between propanone and iodine.
H+
CH3COCH3 + I2 —–> CH3COCH2I + H+ + I-

This reaction is first order with respect to propanone and H+ and zero order with respect to iodine. At a certain temperature, k was found to be
520 mol^1 dm3 s^-1. Calculate the rate at this temperature when
[CH3COCH3] = [I2] = [H+] = 1.50 x 10^-3 moldm^-3

A
  • First use the orders to write the rate equation:
    Rate = k[CH3COCH3]^1 [H+l^1 [I2]^0
  • [X]^1 is usually written as [X]
  • [X]^0 is usually written as 1
  • So you can simplify the equation
    Rate = k[CH3COCH3] [H+]
  • Now, calculate the rate,
    Rate = 520 x [1.50 x 10^-3] x [1.50 x 10^-3]
    Rate = 1.17 x 10^-3 mol dm^-3 s ^-1
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37
Q

What do orders show ?

A

How a reactants concentration affects the rate

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38
Q

What does order 0 show with respect to
[A] ?

A

If [A] changes and the rate stays the same

(So if [A] doubles, the rate will stay the same. If [A] triples the rate will stay the same)

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39
Q

What does order 1 show with respect to [A] ?

A

The rate is proportional to [A]

(So if the rate doubles, the rate will double. If [A] triples the rate will triple)

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40
Q

What does order 2 show with respect to [A] ?

A

The rate is squared.

(So if [A] doubles, the rate will be 4 times faster. If [A] triples the rate will be 9 times faster)

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41
Q

What is an overall order ?

A

This is the sum of the orders of all the different reactants.

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42
Q

What is the general equation for a rate equation ?

A

A + B –> C + D

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43
Q

How do you use experimental data to work out the orders of a reaction ?

A
  • Use the initial rates method to construct a rate-concentration graph and examine the shape
  • Use the initial rates method to directly compare the initial rate for different concentrations
  • Continuously monitor the change in concentration against time and construct a concentration-time graph. Use the graph to compare successive half-lives for the reaction.
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44
Q

What can the shape of the Rate-concentration graph tell you ?

A

The order of the reaction

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45
Q

How do you use the concentration-time graph to construct a rate-concentration graph ?

A
  • Find the gradient, which represents the rate, at various points along the concentration-time graph.
  • On a new grah, plot these rate values against concentration. Join up the points with a line or smooth curve, and you’re done. The shape of the new curve tells you the order.
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46
Q

What is the rate-concentration graph for the order 0 ?

A

A horizontal line

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47
Q

What is the rate-concentration graph for the order 1 ?

A

A straight line that goes through the origin

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48
Q

What is the rate-concentration graph for the order 2 ?

A

A curve

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49
Q

The reaction below was found to be second order with respect to NO and zero order with respect to CO and O. At a certain temperature, the rate is
1.76 x 10^-3 mol dm^-3 s^-1 when
[NO] = [CO] = [O2] = 2.00 x10^-3 mol dm^-3

NO + CO + O2 –> NO2 + CO2

Find out the rate constant at this temperature.

A
  • First write out the rate constant at this temperature.
  • Rate = k[NO]^2 [CO]^0 [O2]0
    Rate = k[NO]^2
  • 1.76 x 10^-3 = k[2.00 x 10^-3]^2
  • k = (1.76 x 10^-3) / [2.00 x 10^-3]2
  • k = 440
  • Find the units of k by putting the units for rate and concentration into the same expression.

Units for k =
mol dm^-3 s^-1/ (mol dm^-3)^2
s^-1 / mol dm^-3
mol^1 dm^3 s^-1

K = 440 mol^1 dm^3 s^-1

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50
Q

What does the order of a reaction with respect to a particular reactant tell you ?

A

How the concentrations affects the rate

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51
Q

If the rate of a reaction doubles which you double the conc. of reactant X (but keep the conc. of all the other reactants and the temperature constant), what is the order of reaction with respect to reactant X ?

A

First order

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52
Q

How do you work out the overall order of a reaction from a rate equation ?

A

The sum of all the orders of all the different reactants.

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53
Q

The ester ethyl ethanoate, CH3COOC2H5, is hydrolysed by heating with dilute acid to give ethanol and ethanoic acid. The reaction is first order with respect to the concentrations of H+ and the ester.

a) Write the equation for the reaction

b) When the initial concentration of the acid is 2.0 mol dm^-3 and the ester 0.25 mol dm^-3, the initial rate is 2.2 x 10^-3 mol dm^-3 s^-1. Find the value of the rate constant at this temperature and give its units.

c) Use your answer from part b) to calculate the initial rate at the same temperature if more solvent is added to the initial mixture so that the volume doubles.

A

a) Rate = k[CH3COOC2H5]^1 [H+]^1

b) 2.2 x 10^-3 = k[0.25] [2.0]
k = (2.2 x 10^-3) / [0.25 x 2.0]
k = 4.4 x 10^-3

Units = mol dm^-3 s^-1 / (mol dm^-3)^2
Units = s^-1 / mol dm^-3
Units = mol^-1 dm^3 s^-1

c) If the volume doubles then the conc. of the reactants halves.
[H+] = 1.0 mol dm^-3
[CH3COOC2H5] = 0.125 mol dm^-3

Rate = (4.4 x 10^-3) x (1.0 x 0.1250
Rate = 5.5 x 10^-4 mol dm^-3 s^-1

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54
Q

What is a half-life ?

A

The time taken for a reactant to halve in quantity

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55
Q

What is the half-life of a reaction ?

A

The time it takes for half of the reactant to be used up.

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56
Q

Can orders be worked out from half lives and concentration-time graphs ?

A

Yes, by looking at the shape of a concentration-time graph and seeing whether the half-life is constant.

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57
Q

How do you know if the graph is zero order ?

A

The rate does not change as concentration falls - the graph is a straight line. The half-life decreases as the reaction goes on.

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58
Q

How do you know if the graph is first order ?

A

The graph is curved. The rate decreases as the conc. does but the half life is constant.

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59
Q

How do you know if the graph is second order ?

A

The graph is curved again, but the half-life increases as the reaction goes on.

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60
Q

Do you need to know the order in respect to the reactants to write a rate equation ?

A

Yes

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61
Q

How do you find the rate constant from the half-life of a first order reaction ?

A
  • The half-life of a first order reaction is independent of the concentration. So each half-life will be the same length
  • This means the half-life of a first order reaction can be read off its concentration-time graph by seeing how long it takes to halve the reactant conc.
  • If you know the half-life of a first order reaction, you can work out the rate constant.
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62
Q

What is the equation when you want to work out the half-life of a first order reaction ?

A

k = In2 / (half-life)

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63
Q

Whats the half-life of a reaction ?

A

The time taken for a reactant to halve in quantity

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64
Q

What are the units for this equation ?

k = In2 / (half-life)

A

s^-1

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65
Q

How can you calculate the half-life of a reaction from a concentration-time
graph ?

A

By looking at the shape of he concentration-time graph and seeing of the half-life is constant.

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66
Q

How can you use the a concentration-time graph to work out the value of k for a first order reaction ?

A
  • The half-life can be read off the concentration-time graph by seeing how long it takes to halve the reactant concentration.

k = In2 / (half-life)

67
Q

[A] Time [B] Time
1 0 1 0
0.5 17 0.5 21
0.25 34 0.25 58
0.125 51 0.125 117

In a reaction between A and B, the concentration of each reactant is measured at regular intervals and concentration-time graphs are plotted for A and B . Neither graph is a straight line. The data in the table is obtained from the concentration-time graphs.

a) Use the data in the table to find the order of the reaction:
i) with respect to A
ii) with respect to B

b) Write a rate equation for the reaction.

A

a) i) For [A] goes from 1 to 0.5 takes 17
For [A] to go from 0.5 to 0.25 takes 17
For [A] to go from 0.25 to 0.125 takes 17

The half lives are constant so the reaction is first order with respect to A.

ii) For [B] goes from 1 to 0.5 takes 21
For [B] goes from 0.5 to 0.25 takes 37
For [B] goes from 0.25 to 0.125 takes 59

The half lifes for reactant B increase, so the reaction is second order with respect to B

b) Rate = k[A][B]^2

68
Q

What affects the rate constant ?

A

Temperature changes.

69
Q

What do the particles need to do, to make a reaction happen ?

A
  • Collide with each other
  • Have enough energy to react
  • Have the right orientation
70
Q

How does increasing the temperature increase the rate ?

A
  • Increasing the temperature gives the reactant particles more kinetic energy. This means the particles speed up, so they collide more often.
  • Increasing the temperature also means more reactant particles will have the required activation energy for the reaction, so a greater proportion of the collisions will result in the reaction happening.
  • In other words, increasing the temperature increases the reaction rate.
71
Q

What is the rate equation ?

A

Rate = k[A]^m [B]^n

72
Q

If changing the temperature affects the rate of reaction, but it doesn’t affect the concentration, what will it affect ?

A

The rate constant

73
Q

If the higher the temperature, will the reaction have a faster or slower rate constant ?

A

The reaction will have a faster rate constant

74
Q

What does the Arrhenius equation link together ?

A

The rate constant and the activation energy and tempurature

75
Q

What is the Arrhenius equation ?

A

k = Ae ^-Ea/RT

Ae = The pre-exponential factor
Ea = Activation energy
R = Gas constant
T = Temperature
k= rate constant

76
Q

As the activation gets bigger, the rate constant … ?

A

Gets smaller

77
Q

If there is a large activation energy, will this mean a fast or a slow rate ?

Why ?

A

Slow rate

If a reaction has a high activation energy, then not many of the reactant particles will have enough energy to react. So only a few of the collisions will result in the reaction actually happening, and the rate will be slow.

78
Q

As the temperature rises, what happens to the rate constant ?

A

The rate constant increases

79
Q

What is the Arrhenius equation, if you put it in logarithmic form ?

A

Ink = -Ea/RT + InA

80
Q

Can you use the Arrhenius equation in logarithmic form to plot graph ?
What are the axis ?

A

Yes

Plotting InK (y-axis) against 1/T (x-axis)

(This will produce a graph with the gradient of -Ea/R and a y-intercept of INA. You can use the graph to find both the activation enthalpy and the pre-exponential factor.)

81
Q

How doe increasing the temperature affect the value of k for a reaction ?

A

As the temperature increases, k increases

82
Q

In the Arrhenius equation, what do terms k, T and R represent ?

A

k - rate constant
T - temperature
R - gas constant

83
Q

How would you find the activation enthalpy of a reaction from a graph of Ink against 1/T ?

A

It will produce a graph with a gradient of
-Ea/R and a y-intercept of InA.
You can use the graph to find both the activation enthalpy, and the pre-exponential factor.

84
Q

The Arrhenius equation is k = Ae^-Ea/RT. Which one of the following answers as true as Ea increases ?
A - k increases and rate of reaction increases
B - k increases and rate of reaction decreases
C - k decreases and rate of reaction increases
D - k decreases and rate of reaction decreases

A

B

85
Q

Which step is the slowest step in a Multi-step reaction ?

A

The rate determining step

86
Q

How many steps can reaction mechanisms have ?

A

One step oor a series of steps

87
Q

Can each step, in a series of steps in a reaction mechanism, have a different
rate ?

A

Yes

88
Q

How is the overall rate determined ?

A

Is decided by the step with the slowest rate - the rate-determining step.

89
Q

What are the rules, to work out if the reactants from the chemical equation are involved in the rate-determining step ?

A
  • If a reactant appears in the rate equation, it must affect the rate. So the reactant, or something derived from it must be in the rate-determining step.
  • If a reactant doesn’t appear in the rate equation, then it isn’t involved in the rate-determining step.
90
Q

Do reactants in the rate equation affect the rate ?

A

Yes

91
Q

What are some important point to remember about rate determining steps and mechanisms ?

A

The rate-determining step doesn’t have to be the first step in the mechanism

The reaction mechanisms can’t usually be predicted from just the chemical equation.

92
Q

Can you predict the rate equation from the rate-determining step ?

A

Yes

93
Q

What does the order of a reaction with respect to a reactant show ?

A

The number of molecules of that reactant which are involved in the rate-determining step.

94
Q

The mechanism for the reaction between chlorine radicals and ozone, consists of two steps:

Cl. + O3 –> ClO. + O2 –> slow (Rate determining step)
ClO. + O3 –> Cl. + 2O2 –> fast

Predicts the rate equation for this reaction.

A

Cl. and O3 must both be in the rate equation, so the rate equation is of the form ; rate = k[Cl.]^m [O3]^n

There’s only one Cl. radical and one O3 molecule in the rate determining step, so the orders, m and n, are both 1.

So the rate equation is:
Rate = k[Cl.][O3]

95
Q

Can you predict the mechanism from the rate equation ?

How ?

A

Yes

Knowing what reactants are in the rate-determining step gives you an idea of the reaction mechanism.

96
Q

What is it meant by a rate determining step ?

A

The slowest step in a multi - step reaction

97
Q

Is the rate-determining step always the first step in the reaction ?

A

No

98
Q

What is the connection between the rate equation and the rate-determining step ?

A

If a reactant appears in the rate equation then it must affect the rate, so it must be in the rate-determining step.

If a reactant doesn’t appear in the rate equation then it isn’t involved in the rate-determining step.

99
Q

The following reaction is first order with respect to H2, and first order with respect to ICl.

H2 + 2ICl –> I2 + 2HCl

a) Write the rate equation for this reaction.

b) The mechanism for this reaction consist of two steps.
i) Identify the molecules that are in the rate-determining step. Justify your answer.

ii) A chemist suggested the following mechanism for the reaction.
2ICl –> I2 + Cl2 slow
H2 + Cl2 –> 2HCl fast

Suggest, with reasons, whether this mechanism is likely to be correct.

A

a) Rate = k[H2][ICl]

b) i) If the molecule is in the rate equation, then it must be in the rate determining step. The reaction is first order with respect to both H2 and ICl. So there will be one molecule of H2, and one molecule of ICl in the rate determining step.

ii) It is likely to be incorrect. H2 and ICl are both in the rate equation, so they must both be in the rate-determining step / the order of the reaction with respect to ICl is 1, so there must be one molecule of ICl in the rate-determining step.

100
Q

Nitrogen dioxide reacts with carbon monoxide in the following reaction.

NO2 + CO –> NO + CO2

The rate equation for this reaction, found by experiments is: Rate = k[NO2]^2.

a) Explain why the rate equation shows that this cannot be a one-step reaction.

b) The mechanism for this reaction has two steps. The rate-determining step is the first one and results in the formation of NO.
Suggest equations for each of the steps of this reaction.

A

a) Since the reaction is zero order with respect to CO, there cannot be any CO molecules in the rate-determining step. But since there is one CO molecule are n the equation, so the reaction must have more steps.

b)
2NO2 –> NO + NO3
N3 + CO –> NO2 + CO2

101
Q

Is it expensive to produce a chemical ?

A

Yes

102
Q

Why does raw materials need to be considered, when finding the most economical way to produce a chemical ?

A

The plant needs to buy chemicals for the reaction - cheap, widely available ones are best.

103
Q

Why does fuel/energy need to be considered, when finding the most economical way to produce a chemical ?

A

Reactions needing high temperature or pressure will use up a lot of energy. Energy is also used in transporting chemicals to, from and around a plant, mixing them and purifying products.

104
Q

Why does overheads/fixed costs need to be considered, when finding the most economical way to produce a chemical ?

A

No matter how much fuel or raw material a company uses, there are certain costs that need to be met regularly. These include staff wages, rent of equipment or space, taxes, insurance, telephone bills, etc.

105
Q

Why does disposal costs need to be considered, when finding the most economical way to produce a chemical ?

A

Any unwanted by-products will have to be disposed of safely - this is subject to government regulations and can be very expensive.

106
Q

Which reactions are normally best, economically ?

A

Reactions with high atom economies and high percentage yields tend to be best because they use fewer raw materials and have fewer waste products. This saves money for a company.

107
Q

What is considered when designing an industrial process ?

A
  • Rate of reaction
  • Product yield
  • Cost
108
Q

What is an advantage to temperature when choosing the conditions for an industrial process ?

A

Reactions go faster at higher temperatures, meaning more product will be made in the same amount of time

109
Q

What is an disadvantage to temperature when choosing the conditions for an industrial process ?

A

High temperatures make reactions more expensive to carry out because of the cost of fuel

110
Q

What is an advantage to Pressure when choosing the conditions for an industrial process ?

A

Higher pressures make gaseous reactions go faster

111
Q

What is an disadvantage to pressure when choosing the conditions for an industrial process ?

A
  • To create high pressure, gas must be pumped into the reaction vessel. Running powerful pumps uses a lot of energy and is expensive.
  • High pressures can be very dangerous. This means that reaction vessel must be made out of a strong material like thick steel, and incorporate safety systems. Again, this is very expensive.
112
Q

What is an advantage to catalysts when choosing the conditions for an industrial process ?

A
  • The right catalyst can make a reaction go quickly at relatively low temperatures - in some cases, no heat is needed at all. This saves money on fuel.
  • Catalysts are a god investment because they don’t get used up.
113
Q

What is an disadvantage to catalysts when choosing the conditions for an industrial process ?

A
  • Industrial catalysts can be expensive.
  • If a catalyst is in the same state as the reactants, it will have to be separated from the reaction mixture once the reaction is complete. This adds an extra step to the industrial process.
114
Q

What is Kc ?

A

The equilibrium constant

115
Q

Are lots of reactions reversible ?

A

Yes

116
Q

What is dynamic equilibrium ?

A

When the rate of the forward reaction is the same as the rate of the reverse reaction.

117
Q

What is the equation for Kc ?

A

Kc = [D]^d [E]^e / [A]^a [B]^b

118
Q

What changes and doesn’t change after
Kc ?

A

Temperature changes after Kc
Pressure doesn’t change

119
Q

If you increase the temperature, of a reaction what happens to the
equilibrium ?

A

The equilibrium shifts in the endothermic direction to absorb the extra heat.

120
Q

If you decrease the temperature, of a reaction what happens to the
equilibrium ?

A

The equilibrium shifts in the exothermic direction to replace the heat.

121
Q

What happens if more or less product is formed ?

A

If more product is formed > Kc will rise

If less product is formed > Kc will decrease

122
Q

If you increase the pressure, of a reaction what happens to the
equilibrium ?

A

It shifts the equilibrium to the side with fewer gas molecules - reduces this pressure.

123
Q

If you decrease the pressure, of a reaction what happens to the
equilibrium ?

A

Shifts the equilibrium to the side with more gas molecules - raising the pressure.

124
Q

What happens to Kc if the pressure changes ?

A

Kc stays the same

125
Q

What effect does a catalyst have on the equilibrium ?

A

No effect and doesn’t affect Kc.

They can’t increase the yield - but they do mean equilibrium is approached faster.

126
Q

What is the Haber process for producing ammonia ?

N2 + 3H2 <–> 2NH3 = -92 kj mol^-1

A
  • This reaction is normally carried out at 400C and 200 atmospheres of pressure
  • High pressure favours the forward reaction, increasing the yield of ammonia. This is because the equilibrium moves to the side with fewer molecules - in this case the product side. High pressure also speeds up the reaction.
  • A pressure of 200 atmospheres is a compromise. A greater yield and a faster rate could be achieved at higher pressures, but maintaining a high pressure is expensive and has the potential to be dangerous.
  • High temperatures make the reaction go faster - it increases the kinetic energy of the reactant molecules. BUT, it also lowers the product yield. This is because the forward reaction is exothermic - high temperatures shift the position of equilibrium to the left, to absorb the extra heat.
  • A reaction temperature of 400C is a compromise. A greater yield could be achieved at lower temperatures, but the reaction would be too slow to be economical - there’s no point waiting years to get to your product, even if you eventually make loads. At higher temperatures the reaction would go faster, but the yield would become too low to produce ammonia efficiently.
127
Q

What are some risks involved in producing a chemical ?

A
  • Some chemicals, especially gases, are highly flammable.
  • Some chemicals are harmful to our health
  • Some chemicals can also damage the environment.
128
Q

Give three examples of the fixed costs that the chemical company has to meet ?

A
  • Raw materials
  • Fuel/Energy
  • Disposal costs
129
Q

Give one advantage and one disadvantage of using high pressure in industry for a gaseous reaction ?

A

Advantage
+ Higher pressures make gaseous reactions go faster

Disadvantages
- To create a high pressure, gas must be pumped into the reaction vessel. Running powerful pumps require a lot of energy and is expensive.

130
Q

When a dynamic equilibrium is reached, what is the relationship between the forward and reverse reactions ?

A

The rate of the forward reaction is the same as the rate of the reverse reaction

131
Q

Will increasing the temperature of an exothermic reaction increase or decrease the amount of product forming ?

A

You increase the amount of product forming

132
Q

How does adding a catalyst to a reversible reaction affect the value of Kc ?

A

They don’t affect the Kc.

133
Q

Give two risks associated with chemical production on an industrial scale

A
  • Some chemicals, are highly flammable
  • Some chemicals also damage the environment.
134
Q

Methanol is produced industrially by reacting carbon monoxide with hydrogen.

CO + 2H2 <–> CH3OH

State, with reason, the effect that increasing the temperature of this reaction would have on
the :

a) the rate,

b) the position of equilibrium.

A

a) The rate will increase because a higher temperature means faster particles that will collide more often, with more energy.

b) Position of the equilibrium will move to the left because the forward reaction is exothermic, so the reverse reaction will be endothermic and the equilibrium will move to try to absorb heat.

135
Q

Two students are discussing the industrial conditions for producing a chemical using a reversible reaction where the forward and reverse reaction is exothermic.

At 200C, the rate of the reaction is fast, but the yield of product is low.
At 50 C, the rate of the reaction is slow, but the yield of the product is high.
A catalyst can be used to increase efficiency of the process.

Ali says that the catalyst will improve the process because it will increase the reaction at 50C.

Bernie says the catalyst will improve the process because it will increase the yield of the reaction at 200C.

Comment on who is correct, and give a reason for your answer.

A

Ali is correct.

Adding a catalyst has no effect on the yield of a substance, but it does increase the rate of the reaction.

136
Q

Calculate Kc, including units, for the reaction:

PCl5 <=> PCl3 + Cl2

At 600 K, the equilibrium concentrations are:
[PCl5] = 0.024 mol dm^-3
[PCl3] = 0.016 mol dm^-3
[Cl2] = 0.016 mol dm ^-3

A

From the expression above, you can see that:
Kc = [PCl3][Cl2] / [PCl5]

So if you put your equilibrium concentrations into the expression, you get;

Kc = [0.016][0.016] / [0.024]
kc = 0.011

Now work out the units for the rate constant by substituting the units for each component into the expression for Kc :

Kc = [mol dm^-3][mol dm^-3] / [mol dm^-3]
Kc = mol dm^-3

So Kc = 0.011 mol dm^-3

137
Q

When the reaction between ethanoic acid and ethanol was allowed to reach equilibrium at 25 C, it was found that the equilibrium mixture contained 2.0 mol dm^-3 ethanoic acid and 3.5 mol dm^-3 ethanol. The Kc of the equilibrium is 4.0 at 25 C. What are the concentrations of the other components ?

CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O

A

Write out the expression for Kc and substitute in all the values you know :

Kc = [CH3COOC2H5][H2O] / [CH3COOH][C2H5OH]

4.0 =[CH3COOC2H5][H2O] / [2.0][3.5]

Rearrange this gives :
[CH3COOC2H5][H2O] = 4.0 x 2.0 x 3.5 = 28.0

But from the equation,
[CH3COOC2H5] = [H2O]

So, [CH3COOC2H5] = [H2O] = 5.3 mol dm^-3

The concentration of CH3COOC2H5 and H2O is 5.3 mol dm^-3

138
Q

What is a diatomic molecule ?

A

A molecule that only contains two atoms bonded to each other.

139
Q

Why is nitrogen a diatomic molecule ?

A

It contains two N atoms

140
Q

What do nitrogen and hydrogen form ?

A

Ammonia
Ammonium ion

141
Q

How is ammonia formed ?

A

The N atom forms covalent bonds by sharing 1 of its valence electrons with each of the 3 H atoms - this leaves 2 electrons as a lone pair on the N atom. Ammonia can form hydrogen bonds with other molecules, which makes it very soluble in water.

142
Q

How can ammonia form dative covalent bonds ?

A

The lone pair of electrons on the N atom. This allows it to act as a ligand forming complex ions with transition metals.

143
Q

What is a ligand ?

A

Ions or neutral molecules that bond to a central metal atom or ion.

144
Q

How is an ammonium ion formed ?

A

The lone pair of electrons is also responsible for ammonia behaving as a base - the molecule forms dative covalent bonds with protons to form the ammoniam ion.

145
Q

What colour is NO ?

A

A colourless gas

146
Q

What colour is N2O ?

A

It has a sweet smell and is a colourless gas

147
Q

What colour is NO2 ?

A

A gas, it is brown, has a sharp odour and is toxic.

148
Q

Can ammonium compounds react with NaOH ?

A

Yes

149
Q

How can you test if a mixture contains ammonium ions ?

A

Add some NaOH to the solution and gently heating the solution. If ammonium ions are present, they will react with the hydroxide ions and ammonia gas will be evolved

150
Q

How can you test for ammonia gas ?

A

Ammonia gas is alkaline, so you can use a damp piece of red litmus paper, and the paper will turn blue if ammonia is present.

151
Q

Can nitrates react with aluminium in the presence of an alkali ?

A

Yes

152
Q

How do you test if there are nitrate ions present ?

A

Warm a solution with sodium hydroxide solution and some aluminium foil or a piece of devarda’s alloy. In the presence of an alkali, the aluminium reduces the nitrate ions to produce ammonia gas. If ammonia gas is given off, it must have contained nitrate ions. If ammonia gas is present the red litmus paper will turn blue.

153
Q

What is the nitrogen cycle ?

A

It shows how you can convert nitrogen into other compounds.

154
Q

What are the relevant half equations ?

A

N2 + 3H2 –> 2NH3
NH3 + H+ –> NH4+
NH4+ + O2 –> NO2- + 4H+ + 2e-
NO2- + H2O –> NO3- + 2H+ + 2e-
2NO3 + 12H+ + 10e- –> N2 + 6H2O
N2 + O2 –> 2NO
N2 + 2O2 –> 2NO2
2N2 + O2 –> 2N2O

155
Q

Describe the bonding in an ammonium ion ?

A

The lone pair of electrons on the N atom means that ammonia can form dative covalent bonds. The dative covalent bonds with protons to form the ammonium ion.

156
Q

Describe a test you could use to determine whether a solution contains ammonium ions ?

A

Add sodium hydroxide to the solution and heat the mixture. If ammonium ions are present, they will react with the hydroxide ions and ammonia gas will be evolved. Use a damp piece of red litmus paper, and it will turn blue.

157
Q

During lightening storms, nitrogen and oxygen in the air can react to form nitrogen monoxide

a) write an equation for this reaction

b) Nitrogen monoxide reacts further with oxygen to form nitrogen dioxide. What would you see during this reaction ?

A

a) N2 + O2 –> 2NO

b) A brown colour would appear

158
Q

A student is carrying out a test to identify the ions in a solution. He adds some sodium hydroxide solution to it along with a piece of Devarda’s alloy, which contains aluminium. He then gently warms the mixture. A gas is evolved that turns a piece of damp red litmus paper blue. Which of the following statements is incorrect ?

a) The gas evolved is an alkali
b) The aluminium acts as an oxidising agent
c) Heating the solution with a piece if aluminium foil would give the same result
d) The solution contains nitrate (V) ions

A

B

159
Q

Can you do experiments to find the equilibrium concentrations ?

A

Yes

160
Q

When would you use calorimetry to calculate the concentrations

A

If one of the substances in your reaction is coloured, it will absorb a particular wavelength of light. You can use a colourimetry to find the absorbance of this wavelength of light at equilibrium, and use a calibration curve to calculate the concentration of the coloured ions.

161
Q

When would you use pH to calculate the concentrations

A

If one side of your reversible reaction contains either an acid or a base, then you can use a pH probe to find the pH of the equilibrium mixture. You can then use the pH calculations to find the concentrations of the acid or base at equilibrium.

162
Q

a) 0.100 mol/dm3 cu2+ ions are mixed with 0.300 mol/dm3 HCl to form the following equilibrium:
Cu2+ + 4Cl- <=> [CuCl4]2-

At equilibrium, the concentrations of [CuCl4]2- is x mol/dm3. Write an expression for Kc in terms of x.

b) At 291K, the conc. of [CuCl4]2- in solution is 0.0637 mol/dm-3. Calculate Kc.

A

a) From the equation, you can see that for every mole of [CuCl4]2- formed, you will lose 1 mole of Cu2+ and 4 moles of Cl- from your initial reactant concentrations.

So if x mol/dm3 of [CuCl4]2- have formed, you will have lost x mol/dm-3 Cu2+ and 4x mol/dm3 Cl-. So the equilibrium concentrations will be
[Cu2+] = (initial conc.) - x
[Cl-] = (initial conc.) - 4x

[Cu2+] = 0.100 - x
[Cl-] = 0.300 - 4x

From your equilibrium conc. you can see that
Kc = [CuCl4]2- / [Cu2+][Cl-]4
Kc = x / [0.100 - x][0.300 - 4x]

b) x = 0.0637 mol/dm-3
Kc = [0.0637] / [0.100 - 0.0637][0.300 - (4 x 0.0637)}
Kc = 4.20 x 10^5 mol-4dm12

163
Q

When 42.5g nitrogen dioxide was heated in a vessel if volume 22.8 dm3 at 500oC, it dissociated to form x mol dm-3 of oxygen in the equilibrium mixture.
2NO2 <=> 2NO + O2

a) Calculate the starting number of moles of nitrogen dioxide

b) Write an expression for Kc in terms of x

c) Calculate the value of Kc at 500oC, and give its units, given that there were 14.1g of oxygen in the equilibrium mixture

A

a) Moles = mass / mr
42.5 / 46 = 0.924 mol

b) Initial [NO2] = n / v
0.0925 / 22.8 = 0.0405 mol/dm3
[NO2] > 0.0405 - 2x
[NO] > 2x
[O2] > x
Kc = [2x]^2[x] / [0.0405 - 2x]^2

c) moles of O2 = m / mr
=14.1 / 32 = 0.441 mol
[O2] = n / v
= 0.441 / 22.8 = 0.0193 mol/dm3
so x = 0.0193
Kc = (2 x 0.0193)^2(0.0193) / (0.0405 - (00193 x 2))^2
= 7.97 mol dm^3

164
Q
A

Chemistry (8 decks)

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