Find the rate of the following reaction: 2 N | vedclass

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Question

(A) The rate of reaction is given by the expression: $	ext{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}

Vedclass · Accepted Answer

$1.3 	imes 10^{-5} \ mol \ dm^{-3} \ s^{-1}$

Vedclass · Answer

(A) The rate of reaction is given by the expression: $	ext{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.Given that the concentration of $NO_2$ increases by $\Delta[NO_2] = 5.2 	imes 10^{-3} \ M$ in $\Delta t = 100 \ s$.The rate of formation of $NO_2$ is $\frac{d[NO_2]}{dt} = \frac{5.2 	imes 10^{-3} \ M}{100 \ s} = 5.2 	imes 10^{-5} \ M \ s^{-1}$.Therefore,the rate of the reaction is $	ext{Rate} = \frac{1}{4} 	imes (5.2 	imes 10^{-5} \ M \ s^{-1}) = 1.3 	imes 10^{-5} \ mol \ dm^{-3} \ s^{-1}$.

Find the rate of the following reaction: $2 \ N_2O_{5(g)} \rightarrow 4 \ NO_{2(g)} + O_{2(g)}$ if the concentration of $NO_2$ increases to $5.2 \times 10^{-3} \ M$ in $100 \ s$.

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