For the reaction 2N_2O_{5(g)} rightleftharpoo | vedclass

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(B) The rate of reaction is given by the expression: $	ext{Rate} = \frac{1}{4} \frac{d[NO_2]}{dt}$.Given that the change in concentration of $NO_2$ i

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

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(B) The rate of reaction is given by the expression: $	ext{Rate} = \frac{1}{4} \frac{d[NO_2]}{dt}$.Given that the change in concentration of $NO_2$ is $\Delta[NO_2] = 2.0 	imes 10^{-2} \, mol \, L^{-1}$ and the time interval is $\Delta t = 5 \, s$.Therefore,$\frac{d[NO_2]}{dt} = \frac{2.0 	imes 10^{-2} \, mol \, L^{-1}}{5 \, s} = 4 	imes 10^{-3} \, mol \, L^{-1} \, s^{-1}$.Substituting this into the rate expression: $	ext{Rate} = \frac{1}{4} 	imes (4 	imes 10^{-3} \, mol \, L^{-1} \, s^{-1}) = 10^{-3} \, mol \, L^{-1} \, s^{-1}$.

Time $(s)$	$0$	$100$	$200$	$300$	$400$	$700$	$800$
$[C_4H_9Cl]$ $(mol \, L^{-1})$	$0.100$	$0.082$	$0.067$	$0.055$	$0.044$	$0.021$	$0.017$

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.

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