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Chemistry Question on Chemical Kinetics

A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as : Ifk1=1.26×10−4s−1,k2=3.8×10−5s−1,If k_{1}=1.26\times10^{-4} s^{-1}, k_{2}=3.8 \times10^{-5} s^{-1}, then the percentage distributions of B and C are respectively

A

80%,20%80\%, 20\%

B

76.83%,23.17%76.83\%, 23.17\%

C

90%,10%90\%, 10\%

D

63.94%,36.06%63.94\%, 36.06 \%

Answer

76.83%,23.17%76.83\%, 23.17\%

Explanation

Solution

For parallel path reaction, kaverage=k1+k2=1.26×10−4+3.8×10−5=1.64×10−4s−1k_{average} =k_{1}+k_{2}=1.26\times10^{-4}+3.8\times10^{-5}=1.64\times10^{-4} s^{-1} Fractional yield of B=k1kaverage=1.26×10−41.64×10−4=0.7683=\frac{k_{1}}{k_{average}}=\frac{1.26\times10^{-4}}{1.64\times10^{-4}}=0.7683 Percentage distribution of B = 76.83% Fractional yield of C=k2kaverage=3.8×10−51.64×10−4=0.2317C=\frac{k_{2}}{k_{average}}=\frac{3.8\times10^{-5}}{1.64\times10^{-4}}=0.2317 Percentage distribution of C = 23.17%