NO_2 required for a reaction is produced by t | vedclass

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(A) The given reaction is $2N_2O_{5(g)} 	o 4NO_{2(g)} + O_{2(g)}$.Rate of reaction is given by: $	ext{Rate} = -\frac{1}{2} \frac{\Delta [N_2O_5]}{\D

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$1.667 	imes 10^{-2}$

Vedclass · Answer

(A) The given reaction is $2N_2O_{5(g)} 	o 4NO_{2(g)} + O_{2(g)}$.Rate of reaction is given by: $	ext{Rate} = -\frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t} = \frac{1}{4} \frac{\Delta [NO_2]}{\Delta t}$.Change in concentration of $N_2O_5$ is $\Delta [N_2O_5] = 2.75 - 3.00 = -0.25 \, mol \, L^{-1}$.Time interval $\Delta t = 30 \, min$.Rate of disappearance of $N_2O_5 = -\frac{\Delta [N_2O_5]}{\Delta t} = -\frac{-0.25}{30} = 8.333 	imes 10^{-3} \, mol \, L^{-1} \, min^{-1}$.From the stoichiometry,$\frac{1}{4} \frac{\Delta [NO_2]}{\Delta t} = -\frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t}$.Therefore,$\frac{\Delta [NO_2]}{\Delta t} = 2 	imes (-\frac{\Delta [N_2O_5]}{\Delta t}) = 2 	imes 8.333 	imes 10^{-3} = 1.667 	imes 10^{-2} \, mol \, L^{-1} \, min^{-1}$.

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