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

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Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $H_2O$?

Similar Questions

For the reaction $2N_2O_{5(g)} \rightleftharpoons 4NO_{2(g)} + O_{2(g)}$ in a closed vessel,the concentration of $NO_2$ increases by $2.0 \times 10^{-2} \, mol \, L^{-1}$ in $5 \, s$. Calculate the rate of reaction.

Write the rate equation for the following reactions:
$1.$ $2 N_2 O_{5(g)} \rightarrow 4 NO_{2(g)} + O_{2(g)}$
$2.$ $2 HI_{(g)} \rightarrow H_{2(g)} + I_{2(g)}$

Assertion : The rate of the reaction is the rate of change of concentration of a reactant or a product.
Reason : Rate of reaction remains constant during the course of reaction.

Why does the rate of any reaction generally decrease during the course of the reaction?

For the reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$,under certain conditions of temperature and partial pressure of the reactants,the rate of formation of $NH_3$ is $0.001 \ kg \ h^{-1}$. The rate of conversion of $H_2$ under the same conditions is:

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