The rate of the reaction 2N_2O_5 to 4NO_2 + O | vedclass

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(C) The rate of reaction $(ROR)$ is defined as:$ROR = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$Given the

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$k' = 2k$ ; $k'' = k/2$

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(C) The rate of reaction $(ROR)$ is defined as:$ROR = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$Given the rate expressions:$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$$\frac{d[NO_2]}{dt} = k'[N_2O_5]$$\frac{d[O_2]}{dt} = k''[N_2O_5]$Substituting these into the $ROR$ expression:$\frac{1}{2} k[N_2O_5] = \frac{1}{4} k'[N_2O_5] = k''[N_2O_5]$Comparing $k$ and $k'$:$\frac{k}{2} = \frac{k'}{4} \implies k' = 2k$Comparing $k$ and $k''$:$\frac{k}{2} = k'' \implies k'' = \frac{k}{2}$

The rate of the reaction $2N_2O_5 \to 4NO_2 + O_2$ can be written in three ways:
$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$\frac{d[NO_2]}{dt} = k'[N_2O_5]$
$\frac{d[O_2]}{dt} = k''[N_2O_5]$
The relationship between $k$ and $k'$ and between $k$ and $k''$ are:

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The rate of the reaction $2N_2O_5 \to 4NO_2 + O_2$ can be written in three ways:$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$$\frac{d[NO_2]}{dt} = k'[N_2O_5]$$\frac{d[O_2]}{dt} = k''[N_2O_5]$The relationship between $k$ and $k'$ and between $k$ and $k''$ are: