Question

The initial concentration of \( N_2O_5 \) in a first order reaction, \( N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g) \), was \( 1.68\times10^{-2}\,\text{mol L}^{-1} \) at \( 310\,K \). The concentration of \( N_2O_5 \) after 10 minutes was \( 0.84\times10^{-2}\,\text{mol L}^{-1} \), what is the rate constant at \( 310\,K \)? ( \( \log 2 = 0.3010 \) )

• A. \(0.0693\,\text{min}^{-1} \)
• B. \(0.693\,\text{min}^{-1} \)
• C. \(6.93\,\text{min}^{-1} \)
• D. \(0.0639\,\text{min}^{-1} \)
• E. \(0.0963\,\text{min}^{-1} \)

Hint:

If concentration halves $\Rightarrow$ use \(\log 2\).

Answer 1

The Correct Option is A

Step 1: Understanding First-Order Kinetics:
A first-order reaction is one where the reaction rate is directly proportional to the concentration of one of the reactants. We can use the integrated rate law to relate concentration, time, and the rate constant.
Step 2: Key Formula or Approach:
The integrated rate law for a first-order reaction is:
\[ k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t} \] where:
\(k\) is the rate constant.
\(t\) is the time elapsed.
\([A]_0\) is the initial concentration of the reactant.
\([A]_t\) is the concentration of the reactant at time \(t\).
An alternative approach is to recognize the concept of half-life.
Step 3: Detailed Calculation:
Method 1: Using the Integrated Rate Law
We are given:
\([A]_0 = [\text{N}_2\text{O}_5]_0 = 1.68 \times 10^{-2}\) mol L\(^{-1}\)
\([A]_t = [\text{N}_2\text{O}_5]_t = 0.84 \times 10^{-2}\) mol L\(^{-1}\)
\(t = 10\) minutes
First, calculate the ratio of concentrations:
\[ \frac{[A]_0}{[A]_t} = \frac{1.68 \times 10^{-2}}{0.84 \times 10^{-2}} = 2 \] Now, substitute the values into the rate law equation:
\[ k = \frac{2.303}{10 \text{ min}} \log_{10}(2) \] Given \(\log_{10}(2) = 0.3010\).
\[ k = \frac{2.303 \times 0.3010}{10} \text{ min}^{-1} \] \[ k = \frac{0.693143}{10} \text{ min}^{-1} \] \[ k \approx 0.0693 \text{ min}^{-1} \] Method 2: Using the Half-Life Concept
Notice that the concentration after 10 minutes (\(0.84 \times 10^{-2}\)) is exactly half of the initial concentration (\(1.68 \times 10^{-2}\)). This means that the half-life (\(t_{1/2}\)) of this reaction is 10 minutes.
For a first-order reaction, the rate constant \(k\) is related to the half-life by the formula:
\[ k = \frac{0.693}{t_{1/2}} \] Substituting \(t_{1/2} = 10\) min:
\[ k = \frac{0.693}{10 \text{ min}} = 0.0693 \text{ min}^{-1} \] Step 4: Final Answer:
The rate constant of the reaction is 0.0693 min\(^{-1}\).