Cyclopropane rearranges to form propene $\mathrm{△}⟶{\mathrm{CH}}_3-\mathrm{CH}={\mathrm{CH}}_2$. This follows first order kinetics. The rate constant is $2.714 \mathrm{x} {10}^{-3}{\sec }^{-}.$ The initial concentration of cyclopropane is $0.29 \mathrm{M}.$ What will be the concentration of cyclopropane after $100 \sec ?$

Rate constant, $\mathrm{k} =2.714\mathrm{x}{10}^{-3} {\sec }^{-},$ 

initial concentration $\mathrm{a} =0.29\mathrm{M};$ 

concentration left after $100 \sec , (\mathrm{a}-\mathrm{x})=?$ 

We know that: rate constant,

                   $\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}.$

Thus $2.714\mathrm{x} {10}^{-3} \sec$ $=\frac{2.303}{100}\log \frac{0.29}{\mathrm{a}-\mathrm{x}};$

                          $$\begin{gathered} \frac{0.27}{2.303}=\log \frac{0.29}{\mathrm{a}-\mathrm{x}} \\ 0.117=\log \frac{0.29}{\mathrm{a}-\mathrm{x}} \end{gathered}$$

Taking antilog, we get Anti $$\begin{gathered} \log 0.117=\frac{0.29}{\mathrm{a}-\mathrm{x}} \\ \mathrm{a}-\mathrm{x}=\frac{0.29}{1.3091}=0.22\mathrm{M} \end{gathered}$$