Chemical Kinetics

(b) Unit of rate of reaction $\frac{d[A]}{d t}=k[A]$

$\text { Unit }=\mathrm{ms}^{-1}$

(a) Rate of disappearance

$\begin{aligned} & A \rightarrow B \\ & -\frac{d[A]}{d t} \Rightarrow \mathrm{ms}^{-1} \end{aligned}$

(d) For 1st order

$\begin{aligned} \frac{-d[A]}{d t} & =k[A] \\ \mathrm{ms}^{-1} & =k \mathrm{~m} \\ k & =\mathrm{s}^{-1} \end{aligned}$

For 2nd order

$\begin{aligned} \frac{-d[A]}{d t} & =k[A]^2 \\ \mathrm{~ms}^{-1} & =k[\mathrm{~m}]^2 \\ k & =\mathrm{m}^{-1} \mathrm{~s}^{-1} \end{aligned}$

So, $k$ depends on order of reaction and unit of $k$ for 1st order reaction is $\mathrm{s}^{-1}$.

For zero order reaction,

$[A]=[A_0]-kt$ ..... (i)

$[A]=\frac{[A_0]}{2}$ (half-life) ..... (ii)

From eqs. (i) and (ii)

$\frac{\left[A_0\right]}{2}=\left[A_0\right]-k t_{1 / 2} \Rightarrow t_{1 / 2}=\frac{[A]_0}{2 k}$

Plot of $t_{1 / 2}$ versus $\left[A_0\right]$ will be straight line.

Initial weight $\left(A_0\right)$ or $a$ is $4 \mathrm{~g}$

Final weight at time $t$ or after reduce $(a-x)$ or $0.2 \mathrm{~g}$

First order reaction; $t=\frac{2.303}{k} \log \left(\frac{a}{a-x}\right)$

$\begin{aligned} t & =\frac{2.303}{2303 \times 10^{-3}} \log \left(\frac{49}{0.29}\right) \\ t & =10^3 \times \log 20 \\ t & =1301 \mathrm{~s} \text { or } t=0.36 \mathrm{~h} \end{aligned}$

∵ Rate $(r)=$ rate constant $(k)$ [conc.] ${ }^{\text {order }}$

$ k=\frac{\text { rate }}{[\text { conc. }]^{\text {order }(=n)}}=\frac{\mathrm{mol} \mathrm{~L}^{-1}}{\left[\mathrm{~mol} \mathrm{~L}^{-1}\right]^n} $

where, $n=$ order of reaction

$ \therefore \quad k=\mathrm{mol}^{1-n}, \mathrm{~L}^{n-1} \mathrm{~s}^{-1} $

For zero order ( $n=0$ )

$ \therefore \quad k=\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} $

Hence, option (a) is the correct answer.

$ \text { Given, In } k=5.0-\frac{12000}{T} $

$ \begin{aligned} &\text { Also, }\\ &\begin{aligned} \because & & \operatorname{In} k & =\log A-\frac{E_a}{R T} \\ & \text { and } & k & =A e^{-E_a / R T} \\ & \text { o, } & \log _e K & =\log _e A=\frac{E_a}{R T} \end{aligned} \end{aligned} $

where, $A=$ pre-exponential constant i.e. frequency factor

$ \begin{aligned} E_a & =\text { activation energy } \\ R & =\text { gas constant }=2 \mathrm{kcal} \mathrm{~mol}^{-1} \\ T_k & =\text { temperature } \\ k & =\text { rate constant } \end{aligned} $

Thus,

$ \begin{array}{rlrl} & \frac{E_a}{R T} =\frac{12000}{T} \\ \text { or } & E_a =12000 \times R \\ \therefore & E_a =\frac{12000 \times 2}{1000}\,\,\,\left(\text { in } \mathrm{kcal} \mathrm{~mol}^{-1}\right) \\ \text { i.e. } & E_a =24 \mathrm{kcal} \mathrm{~mol}^{-1} \end{array} $

Hence, option (a) is the correct answer.

Any reaction of zero order must obey the following equation

$ \begin{aligned} {[A] } & =-k t+[A]_0 \\ \text { or }[A] & =[A]_0+k t \end{aligned} $

This shows that, the concentration of reactant decreases linearly with time, as it is an equation of a straight line $(y=m x+c)$, the plot of $[A]$ versus $t$ will be straight line with slope $=-k$ and intercept on the concentration axis $=[A]_0$ as shown in figure.

Plot of $[A]$ versus $t$ for a reaction of zero order.