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(D) The given reaction is $4 NH_3 + 5 O_2 \rightarrow 4 NO + 6 H_2 O$. According to the rate expression: Rate of reaction $= -\frac{1}{4} \frac{d[NH_3

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$5.4 	imes 10^{-3} \ M/s$

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(D) The given reaction is $4 NH_3 + 5 O_2 ightarrow 4 NO + 6 H_2 O$. According to the rate expression: Rate of reaction $= -\frac{1}{4} \frac{d[NH_3]}{dt} = \frac{1}{6} \frac{d[H_2 O]}{dt}$. Given that the rate of disappearance of $NH_3$ is $-\frac{d[NH_3]}{dt} = 3.6 	imes 10^{-3} \ M/s$. Substituting this into the rate expression: $\frac{1}{4} (3.6 	imes 10^{-3}) = \frac{1}{6} \frac{d[H_2 O]}{dt}$. Therefore,the rate of formation of water is $\frac{d[H_2 O]}{dt} = \frac{6}{4} 	imes 3.6 	imes 10^{-3} \ M/s$. $\frac{d[H_2 O]}{dt} = 1.5 	imes 3.6 	imes 10^{-3} = 5.4 	imes 10^{-3} \ M/s$.

For the reaction $4 NH_3 + 5 O_2 \rightarrow 4 NO + 6 H_2 O$,if the rate of disappearance of $NH_3$ is $3.6 \times 10^{-3} \ M/s$,what is the rate of formation of water?

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