For the reaction 2N_2O_{5(g)} to 4NO_{2(g)} + | vedclass

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Question

(A) The given reaction is $2N_2O_{5(g)} 	o 4NO_{2(g)} + O_{2(g)}$.The rate of reaction is expressed in terms of the rate of change of concentration o

Vedclass · Accepted Answer

$1.3 	imes 10^{-5} \ M \ s^{-1}$

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(A) The given reaction is $2N_2O_{5(g)} 	o 4NO_{2(g)} + O_{2(g)}$.The rate of reaction is expressed in terms of the rate of change of concentration of products as:$	ext{Rate} = \frac{1}{4} \frac{d[NO_2]}{dt}$Given that $\frac{d[NO_2]}{dt} = \frac{5.2 	imes 10^{-3} \ M}{100 \ s} = 5.2 	imes 10^{-5} \ M \ s^{-1}$.Substituting this value into the rate expression:$	ext{Rate} = \frac{1}{4} 	imes (5.2 	imes 10^{-5} \ M \ s^{-1}) = 1.3 	imes 10^{-5} \ M \ s^{-1}$.

For the reaction $2N_2O_{5(g)} \to 4NO_{2(g)} + O_{2(g)}$,if the concentration of $NO_2$ increases by $5.2 \times 10^{-3} \ M$ in $100 \ s$,then the rate of reaction will be:

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For the reaction,$2 \ NO_{(g)} + O_{2_{(g)}} \rightarrow 2 \ NO_{2_{(g)}}$. If $\frac{d[NO_2]}{dt} = 0.052 \ mol \ dm^{-3} \ s^{-1}$,calculate the rate of consumption of $NO_{(g)}$.

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