For the reaction 2N_2O_5(g) rightleftharpoons | vedclass

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(C) The rate of reaction is given by: $Rate = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.Given,the rate o

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$2 	imes 10^{-3}$

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(C) The rate of reaction is given by: $Rate = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.Given,the rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = \frac{2.0 	imes 10^{-2} \ mol \ L^{-1}}{5 \ s} = 4.0 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$.From the rate expression: $-\frac{d[N_2O_5]}{dt} = \frac{1}{2} \frac{d[NO_2]}{dt}$.Substituting the value: $-\frac{d[N_2O_5]}{dt} = \frac{1}{2} 	imes (4.0 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}) = 2.0 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

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 change of concentration of $N_2O_5$.

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$\left[ \frac{-d[N_2 O_5]}{dt} = 0.02 \ mol \ dm^{-3} \ s^{-1} \right]$

$A$ gaseous hypothetical chemical equation $2A \rightleftharpoons 4B + C$ is carried out in a closed vessel. The concentration of $B$ is found to increase by $5 \times 10^{-3} \ mol \ L^{-1}$ in $10 \ s$. The rate of appearance of $B$ is

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