Q16.101 CP - Chemistry: Molecular Nature Of Matter And Change

Question

For the decomposition of gaseous dinitrogen pentaoxide, $\mathbf{2}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{5}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\to }\mathbf{4}\mathbf{N}{\mathbf{O}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{O}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$ , the rate constant is $\mathit{k}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$ at ${\mathbf{60}}^{\mathbf{0}}\mathit{C}$ . The initial concentration of ${\mathit{N}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{5}}$ is 1.58 mol/L. 

(a) What is $\left[N_2{O}_5\right]$ after 5.00 min? 

(b) What fraction of the ${\mathit{N}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{5}}$ has decomposed after 5.00 min?

Step-by-Step Solution

Answer

(a) After 5.00 min, the concentration is equal to 0.68 mol/L.

(b) After 5.00 min, the fraction of decomposed is equal to 0.57.

Step 1: What is [ N 2 O 5 ] after 5.00 min?

The rate constant, k, is expressed in units of $s^{-1}$, implying a first-order reaction. Consider ${\left[N_2{O}_5\right]}_0$ as the starting concentration and ${\left[N_2{O}_5\right]}_t$ as the concentration at time t. For a first-order process, use the integral rate law to determine the reactant concentration:

$$\begin{gathered} \ln \frac{{\left[N_2{O}_5\right]}_t}{{\left[N_2{O}_5\right]}_0}=-kt \\ \ln \frac{{\left[N_2{O}_5\right]}_t}{{\left[1.58mol/L\right]}_0}=-\left({2.8\times 10}^{-3}{s}^{-1}\right)\left(\frac{60s}{1\min }\right) \\ \ln \frac{{\left[N_2{O}_5\right]}_t}{{\left[1.58mol/L\right]}_0}=-0.84 \end{gathered}$$

$$\begin{gathered} \frac{{\left[N_2{O}_5\right]}_t}{{\left[1.58mol/L\right]}_0}=e^{-0.84} \\ \frac{{\left[N_2{O}_5\right]}_t}{{\left[1.58mol/L\right]}_0}=0.43171 \\ {\left[N_2{O}_5\right]}_t=0.43171\times \left(1.58mol/L\right) \\ {\left[N_2{O}_5\right]}_t=0.68mol/L \end{gathered}$$

Thus, the concentration of $N_2{O}_5$ is equal to 0.68 mol/L.

Step 2: What fraction of the N 2 O 5 has decomposed after 5.00 min?

Determine the fraction of $N_2{O}_5$ decomposed:

$$\begin{gathered} Fraction of N_2{O}_5 decomposed=\frac{1.58mol/L-0.68mol/L}{1.58mol/L} \\ =0.57 \end{gathered}$$

Thus, the fraction of decomposed is equal to 0.57.