Question

A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as ; $k_1=1.26\times {10}^{-4}se{c}^{-1}$ and $k_2=3.6\times {10}^{-5}se{c}^{-1}$. Calculate the % distribution of B & C.

• A. 77.7, 22.3
• B. 22.3, 77.7
• C. 67.7, 32.3
• D. None of these

Answer 1

Answer: (A)

77.7, 22.3

Rate law equation for two parallel reactions is,

$Rate=k_1\left[A\right]+k_2\left[A\right]$

$PercentageofB=\frac{k_1}{k_1+k_2}\times 100=\frac{12.6\times {10}^{-5}}{(12.6+3.6)\times {10}^{-5}}=77.7$

$PercentageofC=\frac{k_2}{k_1+k_2}\times 100=\frac{3.6\times {10}^{-5}}{(12.6+3.6)\times {10}^{-5}}=22.3$

Thus the answer should be an option (A).