Chemical Kinetics - Result Question 45
45. For the first order reaction,
$(1998,2 M)$
(a) the degree of dissociation is equal to $\left(1-e^{-k t}\right)$
(b) a plot of reciprocal concentration of the reactant $v s$ time gives a straight line
(c) the time taken for the completion of $75 \%$ reaction is thrice the $\dfrac{1}{2}$ of the reaction
(d) the pre-exponential factor in the Arrhenius equation has the dimension of time, $T^{-1}$
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Answer:
Correct Answer: 45. $(a, d)$
Solution:
- For a first order reaction :
$ \begin{aligned} & \qquad k t=\ln \dfrac{1}{1-\alpha} \quad \text { where, } \alpha=\text { degree of dissociation. } \\ & \Rightarrow \quad 1-\alpha=e^{-k t} \Rightarrow \alpha=1-e^{-k t} \\ & \text { Also } \dfrac{1}{[A]}=\dfrac{e^{k t}}{[A] _0} \text {, i.e. plot of reciprocal of concentration of } \end{aligned} $
reactant $v s$ time will be exponential.
Time for $75 \%=\dfrac{1}{k} \ln \dfrac{100}{100-75}=\dfrac{2 \ln 2}{k}=2\left(t _{1 / 2}\right)$
The Arrhenius equation is :
$ \ln k=\ln A-\dfrac{E _a}{R T} $
The dimensions of $k$ and $A$ must be same. For first order reaction, dimensions of $k$ is $t^{-1}$.
BETA
