Question

Question 6 (18 marks) (a) Given that the rate constant k for the first-order decomposition of compound X is 2.65 x 10-9 s', calculate the percentage of compound X that has decomposed in the first 2250 seconds after the reaction begins. (4 marks) (b) Consider the first order reaction: W2 → 2 Y. If [W2]=0.8 M initially and 0.17 M after 160 seconds, what will [W2] be after 350 seconds? (4 marks) (c) Data for the reaction 3A + 5B → 2C + 3D are given below. Find the rate law and the rate constant for this reaction. Experiment [A] (M) 0.09 0.18 [B] (M) 1.26 1.26 0.42 Initial rate (M/s) 8.10 x 10-4 3.24 x 10-3 3.60 x 10-4 0.18 (7 marks) (d) The decomposition of A B; (g) at 300 °C temperature is a second-order reaction with a rate constant of 0.109 M's'. Calculate the corresponding half-life if 0.0364 mol of A B; (g) is injected into a vessel of 5 cm at 300 °C. (3 marks)

Answer 1

Question 6

(a)

The integrated rate law for a first-order reaction is:

[R] = [R]oe-kt

Where:

[R]t - concentration of reactant remaining at time t

[R]0 - initial concentration of reactant

k - rate constant

t - time

Given:

k = 2.65*10^(-5) s^(-1)

t = 2250 s

Therefore we have:

[R] = [R]oe-kt

[R] = [Roel-2.65x10-5x2250)

[R]+ = [R]. ~ 0.942

[R]+ = 0.942 [R]

The formula to find percentage decomposed is:

$percentage \: \: decomposed=\frac{[R]_0-[R]_t}{[R]_0}\times 100$

Therefore we have:

$percentage \: \: decomposed=\frac{[R]_0-0.942\: [R]_0}{[R]_0}\times 100=5.78\, \: \%$

(b)

Here first we will find the rate constant.

The integrated rate law for a first-order reaction is:

[R] = [R]oe-kt

OR

$k=\frac{1}{t}\times ln\left ( \frac{[R]_0}{[R]_t} \right )$

We have:

[R]t = 0.17 M

[R]0 = 0.8 M

t = 350 s

Thefore value of k is:

$k=\frac{1}{160}\times ln\left ( \frac{0.8}{0.17} \right )=9.68\times 10^{-3}\: s^{-1}$

Now we have to find [R]t at time t = 350 s.

[R] = [R]oe-kt

$[R]_t=0.8\times e^{(-9.68\times 10^{-3}\times 350)}=0.027\, M$

Hence [W2] after 350 seconds is 0.027 M

(c)

Compare experiment 1 and experiment 2:

When going from experiment 1 to experiment 2, [A] has been doubled and [B] remains constant. The rate has increased by a factor of 4. That is, rate becomes four folds when [A] was increased two folds. Therefore the order of reaction with respect to A is two.

Now, compare experiment 3 and experiment 2:

When going from experiment 3 to experiment 2, [B] has been tripled and [A] remains constant. The rate has increased by a factor of 9. That is, the rate becomes nine folds when [B] was increased three folds. Therefore the order of reaction with respect to B is two.

Hence the rate law is:

The constant k can be obtained by using data from experiment 1.

The equation for k is:

$k=\frac{rate}{[A]^2[B]^2}$

Using data from experiment 1, we have:

$k=\frac{8.1\times 10^{-4}}{0.09^2\times 1.26^2}=6.23\times 10^{-2}\: \: mol^{-3}L^3s^{-1}$

(d)

Half-life is the time taken for the reactant concentration to reach half the initial value.

The half-life of a second-order reaction is given by:

$T_{1/2}=\frac{1}{k[R]_0}$

concentration of reactant = number of moles/volume

Here:

number of moles = 0.0364

volume = 5 cm3 = 5 mL = 0.005 L

Therefore:

initial concentration of reactant = 0.0364/0.005 = 7.28 M

k = 0.109 L/mol.s

Hence the half life is:

$T_{1/2}=\frac{1}{k[R]_0}=\frac{1}{0.109\times 7.28}=1.26\: \: s$

Hence the half-life is 1.26 seconds