Question

The rate constant for the second-order reaction $$ \mathrm{NOBr}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) $$ is \(48 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{min}\) at a certain temperature. How long will it take to decompose \(90.0 \%\) of a \(0.0200 \mathrm{M}\) solution of nitrosyl bromide?

Short answer

Answer: The time it takes for 90% of nitrosyl bromide to decompose is approximately 9.375 minutes.

Step-by-step solution

Write down the integrated rate equation for a second-order reaction

For a second-order reaction, the integrated rate equation is: $$ \frac{1}{[A]_t} - \frac{1}{[A]_0} = kt $$ where \([A]_t\) is the concentration at time t, \([A]_0\) is the initial concentration, k is the rate constant, and t is the decomposition time.

Find the concentration after 90% decomposition

When 90% of nitrosyl bromide decomposes, we are left with 10% of the initial concentration. We can calculate this as follows: $$ [A]_t = (1 - 0.9) [A]_0 = 0.1 \times 0.0200 M = 0.00200 M $$

Substitute the values into the integrated rate equation

Now, we will substitute the values of \([A]_t\), \([A]_0\), and k into the integrated rate equation: $$ \frac{1}{0.00200 M} - \frac{1}{0.0200 M} = (48 L/mol \cdot min) \cdot t $$

Solve for the decomposition time

We will now solve the equation for t: $$ \frac{1}{0.00200 M} - \frac{1}{0.0200 M} = (48 L/mol \cdot min) \cdot t \\ 500 M^{-1} - 50 M^{-1} = 48 L/mol \cdot min \cdot t \\ 450 M^{-1} = 48 L/mol \cdot min \cdot t $$ Now, dividing both sides by the rate constant, we get: $$ t = \frac{450 M^{-1}}{48 L/mol \cdot min} = \frac{450 M^{-1}}{48 (mol/L)^{-1} min^{-1}} = 9.375 min $$ The time it takes to decompose 90% of the \(0.0200 M\) nitrosyl bromide solution is approximately 9.375 minutes.