# The decomposition of $N_2O_5$ is given by the following equation: $2\; N_2O_5\; (g) \rightarrow 4 NO_2\;(g) + O_2\;(g)$ At an initial $[N_2O_5]_0 = 3.15\; M$, the initial rate of reaction is $5.45 \times 10^{-5} \;M/s$ and when $[N_2O_5]_0 = 0.78\; M$, the initial rate of reaction is $1.35 \times 10^{-5}\; M/s$. What is the rate constant of this reaction?

$\begin{array}{1 1} 1.73 \times 10^{-5} s^{-1} \\ 3.46 \times 10^{-5} s^{-1} \\ 0.865 \times 10^{-5} s^{-1} \\ 5.12 \times 10^{-5} s^{-1} \end{array}$

Answer: $1.73 \times 10^{-5} s^{-1}$
Given $2\; N_2O_5\; (g) \rightarrow 4 NO_2\;(g) + O_2\;(g)$
We can equate a dimensionless ratio of two rates to a rate law with an as yet an undetermined order, as follows:
$\Rightarrow \large\frac{5.45 \times 10^{-5}} { 1.35 \times 10%{-5}}$$= \large\frac{3.15^a}{0.78^a} \Rightarrow 4.037 = 4.039^a \rightarrow a \approx 1 Now that the order is found to be = 1, we can find the rate constant as \large\frac{5.45\times10^{-5}}{3.15}$$ =1.73 \times 10^{-5} s^{-1}$
answered Jul 24, 2014
edited Jul 24, 2014