For the reaction 4NH_3 + 5O_2 to 4NO + 6H_2O, | vedclass

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(A) The rate of reaction is given by the expression: $-\frac{1}{4} \frac{d[NH_3]}{dt} = +\frac{1}{6} \frac{d[H_2O]}{dt}$.Given that the rate of disapp

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$5.4 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$

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(A) The rate of reaction is given by the expression: $-\frac{1}{4} \frac{d[NH_3]}{dt} = +\frac{1}{6} \frac{d[H_2O]}{dt}$.Given that the rate of disappearance of $NH_3$ is $-\frac{d[NH_3]}{dt} = 3.6 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$.Substituting this value into the expression: $\frac{1}{4} 	imes (3.6 	imes 10^{-3}) = \frac{1}{6} \frac{d[H_2O]}{dt}$.Therefore,the rate of formation of $H_2O$ is $\frac{d[H_2O]}{dt} = \frac{6}{4} 	imes 3.6 	imes 10^{-3} = 1.5 	imes 3.6 	imes 10^{-3} = 5.4 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

For the reaction $4NH_3 + 5O_2 \to 4NO + 6H_2O$,if the rate of disappearance of $NH_3$ is $3.6 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$,what is the rate of formation of $H_2O$?

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