Question

Rate constant \( K \) for a first order reaction is found \( 5.5 \times 10^{-14}\ \text{s}^{-1} \). Calculate the half-life of this reaction.
OR
For the following first order reaction — \( N_2O_5(g) \rightarrow 2NO_2(g) + \tfrac{1}{2}O_2(g) \) — the initial concentration of \( N_2O_5 \) at 318 K was \( 1.24 \times 10^{-2}\ \text{mol L}^{-1} \). After 60 min it became \( 0.20 \times 10^{-2}\ \text{mol L}^{-1} \). Calculate the rate constant at 318 K.

Hint:

For any first order reaction \( t_{1/2} = 0.693/k \); to get \( k \) from concentrations use \( k = \frac{2.303}{t}\log\frac{[A]_0}{[A]} \). Watch the time units (min vs s).

Answer 1

Option 1 (Half-life from rate constant)
Step 1: For a first order reaction the rate does not depend on the initial concentration, and the half-life \( t_{1/2} \) is related to the rate constant \( k \) by the standard formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Here \( 0.693 = \ln 2 \), the natural logarithm of 2.
Step 2: Substitute the given rate constant \( k = 5.5 \times 10^{-14}\ \text{s}^{-1} \):
\[ t_{1/2} = \frac{0.693}{5.5 \times 10^{-14}} \]
Step 3: Carry out the arithmetic. First \( \dfrac{0.693}{5.5} = 0.126 \). Then divide by \( 10^{-14} \), which multiplies by \( 10^{14} \):
\[ t_{1/2} = 0.126 \times 10^{14}\ \text{s} = 1.26 \times 10^{13}\ \text{s} \]
\[\boxed{t_{1/2} = 1.26 \times 10^{13}\ \text{s}}\]

Option 2 (Rate constant from concentration data)
Step 1: For a first order reaction the integrated rate law is:
\[ k = \frac{2.303}{t}\ \log \frac{[A]_0}{[A]} \]
where \( [A]_0 \) is the initial concentration and \( [A] \) the concentration after time \( t \).
Step 2: Insert the given data: \( [A]_0 = 1.24 \times 10^{-2} \), \( [A] = 0.20 \times 10^{-2}\ \text{mol L}^{-1} \), \( t = 60\ \text{min} \).
\[ \frac{[A]_0}{[A]} = \frac{1.24 \times 10^{-2}}{0.20 \times 10^{-2}} = 6.2 \]
Step 3: Take the logarithm: \( \log 6.2 = 0.7924 \).
Step 4: Substitute and compute:
\[ k = \frac{2.303}{60} \times 0.7924 = \frac{1.825}{60} = 0.0304\ \text{min}^{-1} \]
Step 5: Convert to \( \text{s}^{-1} \) by dividing by 60:
\[ k = \frac{0.0304}{60} = 5.07 \times 10^{-4}\ \text{s}^{-1} \]
\[\boxed{k = 0.0304\ \text{min}^{-1} = 5.07 \times 10^{-4}\ \text{s}^{-1}}\]