Rate of equations

Question 1

Rate Equations

Calculation

Answer 1

r = k[A]m[B]n

m, n are called reaction orders

Orders are usually integers 0,1,2 0 means the reaction is zero order with respect to that reactant 1 means first order 2 means second-order

NOTE: the orders are not the same as the stoichiometric coefficients in the balanced equation. They are worked out experimentally.

Question 2

The total order

Answer 2

The total order for a reaction is worked out by adding all the individual orders together (m+n)

Question 3

draw orders graph

describe

Answer 3

Question 4

The rate constant (k)

Answer 4

1. The units of k depend on the overall order of the reaction. It must be worked out from the rate equation
2. The value of k is independent of concentration and time. It is constant at a fixed temperature
3. The value of k refers to a specific temperature and it increases if we increase temperature

Question 5

Continuous Monitoring

Answer 5

When we follow one experiment over time recording the change in concentration we call it a continuous rate method.

The gradient represents the rate of reaction. The reaction is fastest at the start where the gradient is steepest.

The rate drops as the reactants start to get used up and their concentration drops. The graph will eventually become horizontal and the gradient becomes zero which represents the reaction having stopped.

Question 6

Measurement of the change in volume of a gas

Answer 6

Question 7

A large excess of reactants effect on order

Answer 7

In reactions where there are several reactants, if the concentration of one of the reactant is kept in a large excess then that reactant will appear not to affect rate and will be pseudo-zero order . This is because its concentration stays virtually constant and does not affect rate.

Question 8

Comparing continuous rate curves

Answer 8

Question 9

Initial rate method

Answer 9

The initial rate can be calculated from taking the gradient of a continuous monitoring conc vs time graph at time = zero

Initial rate can also be calculated from clock reactions where the time taken to reach a fixed concentration is measured.

Question 10

A Common Clock Reaction

Answer 10

Question 11

Answer 11

Question 12

Answer 12

Question 13

zero order: Calculating k from Concentration-time graphs

Answer 13

For zero order reactants, the rate stays constant as the reactant is used up. This means the concentration of that reactant has no effect on rate. Rate = k [A]0 so rate = k As the rate is the gradient of the graph on the right, the gradient is also the value of the rate constant

Question 14

Effect of Temperature on Rate Constant: The Arrhenius Equation

Answer 14

Increasing the temperature increases the value of the rate constant k

. The relationship is given by the Arrhenius equation k = Ae-EA/RT where A is the Arrhenius constant, R is the gas constant, and EA is the activation energy.

ln k = ln A – EA/(RT)

Question 15

Rate determining step

Question 16

To calculate the initial rate, some assumptions are made:

Answer 16

• The concentration of the reactant does not change significantly over the timescale
• The temperature is constant
• The endpoint has not been over-estimated by observing the colour change or precipitate formed too late

Question 17

By monitoring certain properties of a clock reaction, the rate can be determined. These properties might include:

Answer 17

• Precipitate formation — if the solution is clear and the reaction produces a precipitate which makes it opaque, then the time taken for us to no longer be able to see through the solution is measured

• Colour change — colorimetry is a useful technique if there is a colour change, although if the colour change is very distinct, colorimetry is not needed

Question 18

Iodine Clock Reaction

Answer 18

1. H+ ions and hydrogen peroxide oxidise I– ions to form I 2

H2O2 (aq) + 2I– (aq) + 2H+ (aq)  I2 (aq) + 2H2O (l)

2. A small amount of sodium thiosulphate with a starch indicator is added to give an orange solution

2S2O3 2– (aq) + I2 (aq)  2I– (aq) + S4O6 2– (aq)

Any iodine produced by the first reaction immediately reacts with thiosulphate ions, converting them back to I - ions

Eventually, sodium thiosulphate will be used up and I 2 will accumulate, causing the starch to change colour from orange to dark blue at the endpoint By repeating this reaction with different concentrations of one of the reagents and plotting a graph of rate against concentration, the order of reaction with respect to this reagent can be determined

Question 19

Partial Pressure

Answer 19

The partial pressure of a gas in a mixture is the pressure that the gas would have if it alone occupied the volume occupied by the whole mixture. If a mixture of gases contains 3 different gases then the total pressure will equal the 3 partial pressure added together

P =p1 + p2 + p3

Question 20

Calculate Partial pressure

Answer 20

partial pressure = mole fraction x total pressure

Question 21

mole fraction

Answer 21

mole fraction = number of moles of a gas total number of moles of all gases

Question 22

Example For the following equilibrium

N2 (g) + 3H2 (g ) 2 NH3 (g)

1 mole of N2 and 3 moles of H2 are added together and the mixture is allowed to reach equilibrium. At equilibrium 20% of the N2 has reacted. If the total pressure is 2kPa what is the value of Kp?

Answer 22

Question 23

Effect of changing conditions on the value of Kc or Kp

Answer 23

The larger the Kp the greater the number of products. If Kp is small we say the equilibrium favors the reactants

Kc and Kp only change with temperature. It does not change if pressure or concentration is altered. A catalyst also has no effect on Kc or Kp

Question 24

Effect of temperature on position of equilibrium and Kc

Answer 24

Both the position of equilibrium and the value of Kc or Kp will change it temperature is altered

In this equilibrium which is exothermic in the forward direction If temperature is increased the reaction will shift to oppose the change and move in the backwards endothermic direction. The position of equilibrium shifts left. The value of Kp gets smaller as there are fewer products

Question 25

Effect of pressure on position of equilibrium and Kp

Answer 25

The position of equilibrium will change it pressure is altered but the value of Kp stays constant as Kp only varies with temperature

N2 (g) + 3H2 (g ) —-2 NH3 (g)

In this equilibrium which has fewer moles of gas on the product side If the pressure is increased the reaction will shift to oppose the change and move in the forward direction to the side with fewer moles of gas. The position of equilibrium shifts right. The value of Kp stays the same though as only temperature changes the value of Kp.

Question 26

Increasing pressure

effect on Kp

Answer 26

Increasing pressure does not change Kp. The increased pressure increases the pressure terms on bottom of Kp expression more than the top. The system is now no longer in equilibrium so the equilibrium shifts to the right, increasing mole fractions of products and decreases the mole fractions of reactants. The top of the Kp expression therefore increases and the bottom decreases until the original value of Kp is restored