Rates of Reaction (Rate Equations) Flashcards
A series of experiments are carried out to find the order of reaction with respect to reactant X.
In these experiments, only the concentration of X was changed.
Which graph would show that the reaction is second - order with respect to X?
( Graph with 4 curves, A, B, C and D )
- D
- ( Graph is rate of reaction against concentration of X)
- ( D shows a curved line with an increasing gradient going up )
Solutions of two compounds, W and X, react together in the presence of a soluble catalyst, Y, as shown in the equation
2W + X -> Z
When the concentrations of W, X and Y are all doubled, the rate of reaction increases by a factor of 4.
What is the possible rate equation for this reaction?
- rate = k [ X ] [ Y ]
- ( 2^x = 4 ) ( Reactants are doubled and rate of reaction incrreases by a factor of 4 )
- ( Therefore sum of orders must be 2 )
- ( Since catalyst is always included in the rate equation, another reactant must be included, thus, there being one mole of X, the rate equation must also include X )
The results of an investigation of the reaction between P and Q are shown in this table.
Experiment 1 2 Initial [ P ] / mol dm^-3 1 ) 0.200 2) 0.600 Initial [Q] / mol dm^-3 1 ) 0.500 2) to be calculated Initial rate / mol dm^-3 s^-1 1 ) 0.400 2 ) 0.800
The rate equation is: rate = k [ P ] [ Q ]^2
What is the initial concentration of Q in experiment 2?
- k = 0.4 / [ 0.2 ] [ 0.5 ]^2 ( rate / [ P ] [Q ]^2 )
- 0.8 = k [ 0.6 ] [ x ]^2
- 0.8 = 8 ( 0.6 ) ( x )^2
- x^2 = 0.8 / 8 ( 0.6 ) ( rearrange equation to find x )
- x = square root answer
- x = 0.408
The compound lithium tetrahydridoaluminate ( III ), LiALH4, is a useful reducing agent.
It behaves in a similar fashion to NaBH4.
Carbonyl compounds and carboxylic acids are reduced to alcohols.
However, LiALH4 also reduces water in a violent reaction so that it must be used in an organic solvent.
Which one of the following concerning the violent reaction between LiALH4 and water is false?
A ) A gas is produced
B ) The activation energy for the reaction is relatively high
C ) The reaction has a negative free - energy change
D ) Aqueous lithium ions are formed
- B
This question involves the use of kinetic data to deduce the order of a reaction and calculate a value for a rate constant.
The data in Table 1 were obtained in a series of experiments on the rate of the reaction
between compounds A and B at a constant temperature.
Experiment 1 ) 2 ) 3 ) Initial concentration of A / mol dm^-3 1 ) 0.12 2 ) 0.36 3 ) 0.72 Initial concentration of B / mol dm^-3 1 ) 0.26 2 ) 0.26 3 ) 0.13 Initial rate / mol dm^-3 s^-1 1 ) 2.10 x 10^-4 2 ) 1.89 x 10^-3 3 ) 3.78 x 10^-3
Show how these data can be used to deduce the rate expression for the reaction between A and B.
- Consider experiments 1 and 2, where B is constant
- [ A ] increases by x3, rate increases by 3^2
- Therefore A is second order
- Consider experiments 2 and 3
- [ A ] increases by x2 so rate should increase by 2^2
- However the rate only increases by 2
- Therefore halving [ B ] halves the rate
- So B is first order
- Rate = k [ A ]^2 [ B ]
The data in Table 2 were obtained in two experiments on the rate of the reaction
between compounds C and D at a constant temperature.
Experiment 4 ) 5 ) Initial concentration of C / mol dm^-3 4 ) 1.9 x 10^-2 5 ) 3.6 x 10^-2 Initial concentration of D / mol dm^-3 4 ) 3.5 x 10^-2 5 ) 5.4 x 10^-2 Initial rate / mol dm^-3 s^-1 4 ) 7.2 x 10^-4 5 ) To be calculated
The rate equation for this reaction is
rate = k[ C ]^2 [ D ]
Use the data from experiment 4 to calculate a value for the rate constant, k, at this
temperature. Deduce the units of k.
( Using expt 4 )
- 7.2 x 10^-4 = k ( 1.9 x 10^-2 )^2 ( 3.5 x 10^-2 )
- k = ( 7.2 x 10^-4 ) / ( 1.9 x 10^-2 )^2 ( 3.5 x 10^-2 )
- k = 56.98456668
- k = 57 mol^-2 dm^6 s^-1
Calculate a value for the initial rate in experiment 5.
k = 57
( rate = k [ C ]^2 [ D ] )
( [ C ] = 3.6 x 10^-2 )
( [ D ] = 5.4 x 10^-2 )
( Using experiment 5 )
x = 57 ( 3.6 x 10^-2 )^2 ( 5.4 x 10^-2 ) ( rate = [ C ]^2 [ D ] )
rate = 3.99 x 10^-3 mol dm^-3 s^-1
The rate equation for a reaction is
rate = k[ E ]
Explain qualitatively why doubling the temperature has a much greater effect on the
rate of the reaction than doubling the concentration of E.
- Reaction occurs when energy is greater than the activation energy
- Doubling the temperature causes many more molecules to have this energy
- However doubling the concentration of E only doubles the number of molecules with the same amount of energy
A slow reaction has a rate constant k = 6.51 x 10^-3 mol^-1 dm^3
at 300 K.
Use the equation ln k = ln A - Ea / RT to calculate a value, in kJ mol^-1, for the
activation energy of this reaction.
The constant A = 2.57 × 10^10 mol^-1 dm^3.
The gas constant R = 8.31 J K^-1 mol^-1.
- ln ( 6.51 x 10^-3 ) = ln ( 2.57 × 10^10 ) - Ea / ( 8.31 ) ( 300 )
- Ea = ( 8.31 ) ( 300 ) ( ln ( 2.57 × 10^10 ) - ln ( 6.51 x 10^-3 ) ) / 1000 ( Ea = RT ( ln A - ln k ) / 1000 ( to get to kj )
- Ea = 72.3 kj mol^-1
Butadiene dimerises according to the equation.
2C4H6 -> C8H12
The kinetics of the dimerisation are studied and the graph of the concentration of a
sample of butadiene is plotted against time. The graph is shown below.
( Graph shows a gradient decreasing at a decreasing rate )
( Graph starts at 0.018 in concentration of Butadiene )
( Graph ends around 0.0056 in concentration of Butandiene )
The initial rate of reaction in this experiment has the value 4.57 × 10^-6 mol dm^-3 s^-1.
Use this value, together with a rate obtained from your tangent, to justify that the
order of the reaction is 2 with respect to butadiene.
( gradient from tangent = -0.0160 - 0 / 7800 )
( = -2.05 x 10^-6 )
- Gradient from tangent = - 0.0160 - 0 / 7800 - 0
- Gradient = = - 2.05 × 10^-6
- Rate = 2.05 × 10^-6 ( mol dm^-3 s^-1 )
Comparison of rate:
- 4.57 × 10^-6 / 2.05 × 10^-6
- ( Initial rate / rate at 0.0120 )
- = 2.23
- 0.018 / 0.012 = 1.5
- ( Initial concentration / concentration at point where tangent drawn )
Deduction of order:
- ( 1.5 )^2 = 2.25
- ( second order is the square of the concentration = rate )
- this is approximately equal to 2.23 therefore order is 2nd with respect to butadiene
The rate of hydrolysis of an ester X (HCOOCH2CH2CH3) was studied in alkaline
conditions at a given temperature. The rate was found to be first order with respect to the
ester and first order with respect to hydroxide ions.
Name ester X.
- propyl methanoate
Using X to represent the ester, write a rate equation for this hydrolysis reaction.
- rate = k[ X ] [ OH- ]
In a second experiment at the same temperature, water was added to the
original reaction mixture so that the total volume was doubled.
Calculate the initial rate of reaction in this second experiment.
( 8.5 x 10^-5 is the initial rate of reaction )
( Total amount of order = 2 )
- 2.13 × 10^-5
- 8.5 x 10^-5 / 4 = 2.125 x 10^-5 ( 2^2 = 4 )
( total amount of order is = 2 )
( since volume is doubled, 2^ of the order )
State the effect, if any, on the value of the rate constant k when the temperature is lowered but all other conditions are kept constant.
Explain your answer.
- Effect: Lowered
- Explanation: fewer particles have enough activation energy to react
Compound A reacts with compound B as shown by the overall equation.
A + 3B -> AB3
The rate equation for the reaction is
rate = k [ A ] [ B ]^2
A suggested mechanism for the reaction is
Step 1 A + B -> AB
Step 2 AB + B -> AB2
Step 3 AB2 + B -> AB3
Deduce which one of the three steps is the rate-determining step.
Explain your answer.
- Rate-determining step: Step 2
- Explanation: involves one mole
A and two Bs
In the presence of the catalyst rhodium, the reaction between NO and H2 occurs according to the following equation.
2NO(g) + 2H2(g) N2(g) + 2H2O(g)
The kinetics of the reaction were investigated and the rate equation was found to be
rate = k [ NO ]^2 [ H2 ]
Using the rate equation and the overall equation, the following three-step mechanism for the reaction was suggested. X and Y are intermediate species.
Step 1 NO + NO -> X
Step 2 X + H2 -> Y
Step 3 Y + H2 -> N2 + 2H2O
Suggest which one of the three steps is the rate-determining step.
Explain your answer.
- Rate-determining step: Step 2
- Explanation: One H2 and two NO appear in rate equation
The compound (CH3)3CBr reacts with aqueous sodium hydroxide as shown in the following equation.
(CH3)3CBr + OH- -> (CH3)3COH + Br-
This reaction was found to be first order with respect to (CH3)3CBr but zero order
with respect to hydroxide ions.
The following two-step process was suggested.
Step 1 (CH3)3CBr -> (CH3)3C+ + Br-
Step 2 (CH3)3C+ + OH- -> (CH3)3COH
Deduce the rate-determining step in this two-step process.
- Step 1
Outline a mechanism for this step using a curly arrow.
Question with equation:
(CH3)3CBr -> (CH3)3C+ + Br−
- Displayed formula, arrow is going from the bond between C-Br to the Br molecule
- ( No charges are present within the displayed formula )
