A substance undergoes first order decomposition involving two parallel first order reaction as:
$K_{1} = 1.25 \times 10^{- 4} S^{- 1}$
$K_{2} = 3.80 \times 10^{- 5} S^{- 1}$
The mole percent of $B$ in the products is:

23.17

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76.68

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30.16

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69.84

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Solution

The correct option is B 76.68
For given parallel $1^{s t}$ order reaction,
mole % of $B = \frac{K_{1}}{K_{1} + K_{2}} \times 100$
$= \frac{1.25 \times 10^{- 4}}{1.25 \times 10^{- 4} + 0.38 \times 10^{- 4}} \times 100$
$= 76.68$ %

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Similar questions

B

k_{1} = 1.26 \times 10^{- 4} s e c^{- 1}

k_{2} = 3.6 \times 10^{- 5} s e c^{- 1}

A substance undergoes first order decomposition involving two parallel first order reaction as :

The mol percent of B in the product is :

K_{1} = 1.26 \times 10^{- 4} s e c^{- 1}

K_{2} = 3.8 \times 10^{- 5} s e c^{- 1}

K_{1} : K_{2} = 1 : 2

B

C

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