Answer the following questions on the above curve for a first order reaction A → P.
(a) What is the relation between slope of this line and rate constant.
(b) (i) Calculate the rate constant of the above reaction if the slope is 2 x 10^–4 s^–1.
     (ii) Derive the relationship between half life of a first order and its rate constant.

(a) Slope = – k

or k = – 2 x 10^–4 mol^–1 Ls^–1.

(ii) Consider the following first order reaction
A → P

at t = 0 a 0

at t = t (a – x) x

Suppose a is the initial concentration of reactant. After time t, x gm mol lit^–1 is changed to P. According to law of mass action the rate of reaction at time t is directly proportional to the concentration of A at that instant i.e., (a – x)

Hence $\frac{-\mathrm{dx}}{\mathrm{dt}}\mathrm{\alpha }(\mathrm{a}-\mathrm{x}) \mathrm{or} \frac{-\mathrm{dx}}{\mathrm{dt}}=\mathrm{x}(\mathrm{a}-\mathrm{x}) \mathrm{where}$

is the rate and K is the rate constant, after $\frac{\mathrm{dx}}{\mathrm{dt}}$ rearranging the above equation, we get

$\frac{-\mathrm{dx}}{(\mathrm{a}-\mathrm{x})}=\mathrm{k} \mathrm{dt}$        ...(i)
On integrating the above equation

${\int }_{}^{}-\frac{\mathrm{dx}}{(\mathrm{a}-\mathrm{x})} = {\int }_{}^{}\mathrm{K} \mathrm{dt}$

We get
               $$\begin{gathered} -\mathrm{In}(\mathrm{a}-\mathrm{x}) = \mathrm{kt} + \mathrm{constant} \\ \left[\therefore {\int }_{}^{}\frac{\mathrm{dx}}{\mathrm{a}-\mathrm{x}}=\ln (\mathrm{a}-\mathrm{x})\right] ...\left(\mathrm{ii}\right) \end{gathered}$$

When t = 0, x = 0 hence – In a = constant substituting this value of constant in eqn. (ii) we get

$$\begin{gathered} -\mathrm{In}(\mathrm{a}-\mathrm{x}) = \mathrm{kt}-\mathrm{In} \mathrm{a} \\ \mathrm{or} \mathrm{In} \mathrm{a} - \mathrm{In}(\mathrm{a}-\mathrm{x}) = \mathrm{kt} \\ \mathrm{or} \mathrm{In}\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} = \mathrm{kt} \\ \left[\because \mathrm{In} \mathrm{A} - \mathrm{In} \mathrm{B} = \mathrm{In}\frac{\mathrm{A}}{\mathrm{B}}\right] \\ \mathrm{or} \mathrm{k} = \frac{1}{\mathrm{t}}\mathrm{In} \frac{\mathrm{a}}{\mathrm{x}-\mathrm{x}} \\ \mathrm{or} \mathrm{k} = \frac{2.303}{\mathrm{t}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}. \end{gathered}$$