Q.For the reaction $N_2O_5(g) \rightarrow 2NO_2(g) + \dfrac{1}{2}O_2(g)$, the value of rate of disappearance of $N_2O_5$ is given as $6.5\times10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}$. The rate of formation of $NO_2$ and $O_2$ is given respectively as

Step 1. For $N_2O_5(g)\rightarrow2NO_2(g)+\tfrac{1}{2}O_2(g)$, $\text{Rate}=-\dfrac{d[N_2O_5]}{dt}=\dfrac{1}{2}\dfrac{d[NO_2]}{dt}=2\dfrac{d[O_2]}{dt}$.

Step 2. Given $-\dfrac{d[N_2O_5]}{dt}=6.5\times10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}$: from $\dfrac{1}{2}\dfrac{d[NO_2]}{dt}=6.5\times10^{-2}$, $\dfrac{d[NO_2]}{dt}=2\times6.5\times10^{-2}=1.3\times10^{-1}\ \text{mol L}^{-1}\text{s}^{-1}$. …

Write the single reaction rate using each species' own stoichiometric coefficient (2 for $NO_2$, 1/2 for $O_2$), then solve for each species' ra …

• Mixing up which species gets multiplied and which gets divided -- $NO_2$'s larger coefficient (2) means it forms FASTER than $N_2O_5$ disappears, …