For the reaction, $C_{3} H_{8} + 5 O_{2} ⟶ 3 {CO}_{2} + 4 H_{2} O$ , how does the rate of disappearance of $O_{2}$ compare to the rate of production of ${CO}_{2}$ ? The rate of disappearance of $O_{2}$ is :

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$\frac{3}{5}$ the rate of production of ${CO}_{2}$ .

$\frac{1}{5}$ the rate of production of ${CO}_{2}$ .

$\frac{5}{3}$ the rate of production of ${CO}_{2}$ .

the same as the rate of production of ${CO}_{2}$ .

answer is D.

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Detailed Solution

Here,

$\frac{- d [O_{2}]}{d t} = + \frac{5}{3} \frac{d [{CO}_{2}]}{d t}$

For the reaction, the rate of disappearance of $O_{2}$ is $\frac{5}{3}$ the rate of production of ${CO}_{2}$ .

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