NO_2 required for a reaction is produced by d | vedclass

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(B) The rate of reaction $(ROR)$ is given by: $ROR = -\frac{1}{2} \frac{\Delta [N_2 O_5]}{\Delta t} = \frac{1}{4} \frac{\Delta [NO_2]}{\Delta t}$.Firs

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(B) The rate of reaction $(ROR)$ is given by: $ROR = -\frac{1}{2} \frac{\Delta [N_2 O_5]}{\Delta t} = \frac{1}{4} \frac{\Delta [NO_2]}{\Delta t}$.First,calculate the rate of disappearance of $N_2 O_5$: $-\frac{\Delta [N_2 O_5]}{\Delta t} = -\frac{(2.75 - 3.00)}{30} = \frac{0.25}{30} = 8.33 	imes 10^{-3} \, mol \, L^{-1} \, min^{-1}$.Using the stoichiometric relationship: $\frac{1}{4} \frac{\Delta [NO_2]}{\Delta t} = -\frac{1}{2} \frac{\Delta [N_2 O_5]}{\Delta t}$.Therefore,$\frac{\Delta [NO_2]}{\Delta t} = 2 	imes (-\frac{\Delta [N_2 O_5]}{\Delta t}) = 2 	imes \frac{0.25}{30} = \frac{0.5}{30} = 0.01666 \, mol \, L^{-1} \, min^{-1}$.Converting to the form $x 	imes 10^{-3}$: $0.01666 = 16.66 	imes 10^{-3} \, mol \, L^{-1} \, min^{-1}$.Rounding to the nearest integer,$x \approx 17$.

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