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Chemistry Question on Chemical Kinetics

The first order integrated rate equation is

A

k=xtk = \frac{x}{t}

B

k=2.303tlogaa−xk = \frac{2.303}{t} log \frac{a}{a-x}

C

k=1tlnaa−xk = \frac{1}{t} ln \frac{a}{a-x}

D

k=1taa(a−x)k = \frac{1}{t} \frac{a}{a\left(a-x\right)}

Answer

k=1tlnaa−xk = \frac{1}{t} ln \frac{a}{a-x}

Explanation

Solution

For first order, rate =d[R]dt=k[R]= \frac{d[R]}{dt} = k[R] or d[R][R]=k dt...(i)\frac{d[R]}{[R]} = k \, dt ... (i) On integrating E (i) ∫d[R][R]=k ∫dt\int \frac{d[R]}{[R]} = k \, \int dt ln [R] = -kt + C \hspace25mm ... (ii) At t = O,[R] = [R0][R_0] [where, R = final concentration, i.e., a - x and R0R_0 is the initial concentration, i.e., a.] ln [R0]=Cln \, [R_0] = C On putting the value of C in E (ii), we get ln [R]=−kt+ln [R0]ln \, [R] = -kt + ln \, [R_0] −kt=ln [R]−ln [R0]-kt = ln \, [R] - ln \, [R_0] kt=ln [R0]−ln [R]kt = ln \, [R_0] - ln \, [R] or k=1tln [R0][R]k = \frac{1}{t} ln \, \frac{[R_0]}{[R]} or k=1tln aa−xk = \frac{1}{t} ln \, \frac{a}{a-x}