The reaction between 2-bromo-2-methylpropane (CH3CBr)3, and warm aqueous 0.100
mol dm−3 sodium hydroxide solution to form 2-methylpropan-2-ol, is found to
have the following rate expression.
rate =k[(CH3)3CBr]
(a) (i)
State what is meant by the term rate of reaction.
(1)
(ii) State the units for the rate constant, k.
(1)
(iii) Deduce what the effect on the rate of the reaction will be if the concentration
of the sodium hydroxide solution is increased to 0.300 mol dm−3.
(1)
(iv) Deduce the overall order of the reaction.
(1)
(v) It is found that after 32.0 seconds the concentration of 2-bromo-2methylpropane falls to 161 of its initial value. Calculate the half-life for the
reaction.
(1)
(Total 5 marks)
(i)
the increase in the concentration of a product / decrease in the
concentration of a reactant per unit time
1
(ii)
s–1/ min–1 etc.
1
(iii)
no effect (as [OH–] does not appear in the rate expression)
1
(iv)
first order / one
1
(v)
8.0 seconds (since four half-lives elapse)
1
(a) The following data were obtained for the reaction of nitrogen
monoxide gas, NO(g), with oxygen gas to form nitrogen dioxide gas,
NO2(g), at 25 C.
Initial rate
/mol dm−3
s−1
Experimen
t
[NO]/
mol dm−3
[O2]/
mol dm−3
1
0.50
0.20
2
0.50
0.40
3.0 × 10−3
6.0 × 10−3
3
1.00
0.80
4.8 × 10−2
(i)
Calculate the order with respect to the two reactants and
write the rate expression for the reaction. Show your
reasoning. (5)
(ii) Explain why the following mechanism is not consistent
with the rate expression.
NO (g) + O2 (g) → NO2(g) + O (g); slow step
NO (g) + O (g) → NO2 (g); fast step
(2)
(iii) Explain why the following mechanism is consistent with
the rate expression, but is unlikely.
2NO(g) +O2 (g) → 2NO2 (g) (2)
(iv) Explain why the following mechanism is consistent with
the rate expression.
NO (g) + O2(g) → NO3 (g); fast
NO3 (g) + NO (g) → 2NO2 (g); slow
(3)
(Total 12 marks)
(i)
[NO] constant, [O2] doubles, rate doubles;
rate  [O2]/ first order;
[NO] doubles, [O2] doubles, rate increases 8 times;
rate  [NO]2/ second order;
rate = k [NO]2[O2] / rate  [NO2]2[O2];
5
(ii)
(iii)
(iv)
slow step / rate determining step involves only one NO and one O2;
not two NO and one O2 (as required by rate-expression);
2
since two NO and one O2 involved in the (one step) mechanism,
correct O2 rate expression possible; but unlikely that three particles
will collide at the same time;
2
from fast step depends on [NO3] depends on [NO] and [O2];
rate depends on [NO3] and [NO];
thus rate must depend on [NO]2[NO2] (consistent with rate expression);
3
The following reaction
2N2O5(g) → 4NO2(g) + O2 (g)
is described as first order with respect to N2O5.
(a) Write the rate expression for the reaction.
(1)
(b) One possible mechanism for this reaction is given below.
N2O5(g) → NO(g) + NO3(g)
Step 1
N2O5(g) + NO3(g) → 2NO2(g) + O2(g) Step 2
Describe the rate expression that would result if the rate
determining step in the mechanism is
(i)
Step 1.
(1)
(ii) Step 2.
(2)
Outline your reasoning.
(c) Explain what is meant by the term half-life for this reaction.
(1)
(d) State what is characteristic about the half-life of a first order
reaction.
(1)
(Total 6 marks)
(a)
rate = k [N2O5]/ k [N2O5]1;
(b)
(i)
(ii)
(c)
1
Rate ×[N2O5]
One molecule of [N2O5];
1
Rate ×[N2O5] [NO3]
But NO3 comes from N2O5 in first step
So:Rate ×[N2O5]2
2
the time taken for the concentration/amount (of N2O5) to decrease to
half its original value / OWTTE;
1
(d)
it is constant / does not depend on concentration (of reactant);
1