A gaseous hypothetical chemical equation $2 A ⇌ 4 B + C$ is carried out in a closed vessel. The concentration of C is found to increase by $5 \times 10^{- 3} mol L^{- 1}$ in 10 second. The rate of appearance of C is

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$4 \times 10^{- 4} {mol L}^{- 1} {sec}^{- 1}$

$5 \times 10^{- 4} {mol L}^{- 1} {sec}^{- 1}$

$5 \times 10^{- 5} {mol L}^{- 1} {sec}^{- 1}$

$6 \times 10^{- 5} {mol L}^{- 1} {sec}^{- 1}$

answer is A.

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Detailed Solution

Increase in concentration of $C = 5 \times 10^{- 3} mol L^{- 1}$

Time = 10 sec

Rate of appearance of C = $\frac{5 \times 10^{- 3}}{10}$ = 5 $\times$ 10^-4 mole/lit sec

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