Question

(a) What is rate of reaction? Write two factors that affect the rate of reaction.

(b) The rate constant of a first order reaction increases from $4\times {10}^{-2}to8\times {10}^{-2}$ when the temperature changes from ${27}^{\circ }C$ to ${37}^{\circ }C$. Calculate the energy of activation $\left(E_a\right)$. (log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021

Answer 1

(a) Rate of reaction may be defined as the change in any one of the reactants or products per unit time.

Concentration of reactants, temperature and catalyst affect the rate of reaction.

(b) Given :

$k_1=4\times {10}^{-2}{k}_2=8\times {10}^{-2}{T}_1=300K{T}_2=310K$

Solution:

$log\left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

$log\left(\frac{8\times {10}^{-2}}{4\times {10}^{-2}}\right)=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

$0.301=\frac{E_a}{2.303\times 8.314J{K}^{-1}mo{l}^{-1}}\left[\frac{310-300}{310\times 300}\right]$

$E_a=\frac{0.301\times 2.303\times 8.314\times 93000}{10}$

$E_a=53598.5J$

OR

(a) (i) For a reaction, A + B $\to$ P

$Rat{e}_1=k\left[A\right][B{]}^2$

If the concentration of B is doubled,

$Rat{e}_2=k\left[A\right][2B{]}^2$

$\frac{Rat{e}_1}{Rat{e}_2}=\frac{k\left[A\right][2B{]}^2}{k\left[A\right][2B{]}^2}$

$Rat{e}_2=k\left[A\right][2B{]}^2$

$\frac{Rat{e}_1}{Rat{e}_2}=\frac{B^2}{4B^2}$

$Rat{e}_2=4Rat{e}_1$

The rate of reaction will be four times the initial rate.

(ii) If A is present in large excess, then the rate of the reaction will be independent of A and will depend only on the concentration of B. The overall rate of the reaction will be 2.

(b) Solution :

Let us first calculate k.

$t_{\frac{1}{2}}=23.1min$

$t_{\frac{1}{2}}=\frac{0.693}{k}k=\frac{0.693}{23.1}$

k = 0.03

Now let us calculate the time required to complete $75\%$ reaction.

$k=\frac{2.303}{t}log\frac{\left[R_0\right]}{\left[R\right]}$

$0.03=\frac{2.303}{t}log\frac{100}{2.5}0.03=\frac{2.303}{t}log4t=\frac{2.303}{0.03}log4$

t = 46.22 mins