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For the reaction,
$2 C_{2} H_{6} (g) + 7 O_{2} (g) ⟶ 4 {CO}_{2} (g) + 6 H_{2} O (l)$
the rate of disappearance of $C_{2} H_{6} (g) :$

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equals the rate of disappearance of $O_{2} (g)$ .

is seven times the rate of disappearance of $O_{2} (g)$

is one-third the rate of appearance of $H_{2} O (l)$

is twice the rate of appearance of ${CO}_{2} (g)$

answer is D.

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Detailed Solution

$- \frac{d [C_{2} H_{6}]}{d t} = - \frac{2}{7} \frac{d [O_{2}]}{d t} = \frac{1}{2} \frac{d [{CO}_{2}]}{d t} = \frac{1}{3} \frac{d [H_{2} O]}{d t}$

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