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 using \(\ln 2\))
Step 1: A first order decay obeys \([A] = [A]_0 e^{-kt}\). Setting \([A] = \tfrac{1}{2}[A]_0\) gives \(\tfrac{1}{2} = e^{-k t_{1/2}}\).
Step 2: Taking natural logs, \(k\, t_{1/2} = \ln 2\), so \(t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{k}\). This shows the half-life is fixed and independent of how much reactant you start with.
Step 3: Plug in \(k = 5.5 \times 10^{-14}\ \text{s}^{-1}\): \(t_{1/2} = \dfrac{0.693}{5.5 \times 10^{-14}}\).
Step 4: Evaluate: \(0.693 \div 5.5 = 0.126\), and dividing by \(10^{-14}\) raises the power to \(10^{13}\).
\[\boxed{t_{1/2} = 1.26 \times 10^{13}\ \text{s} \approx 4.0 \times 10^{5}\ \text{years}}\]

Option 2 (Rate constant via natural logarithm)
Step 1: Use the natural-log form of the first order law: \(\ln \dfrac{[A]_0}{[A]} = k t\), hence \(k = \dfrac{1}{t}\ln \dfrac{[A]_0}{[A]}\).
Step 2: The concentration falls from \(1.24 \times 10^{-2}\) to \(0.20 \times 10^{-2}\ \text{mol L}^{-1}\), a ratio of \(6.2\), over \(t = 60\ \text{min}\).
Step 3: \(\ln 6.2 = 1.825\).
Step 4: Therefore \(k = \dfrac{1.825}{60\ \text{min}} = 0.0304\ \text{min}^{-1}\).
Step 5: Since \(1\ \text{min} = 60\ \text{s}\), \(k = \dfrac{0.0304}{60} = 5.07 \times 10^{-4}\ \text{s}^{-1}\).
\[\boxed{k = 0.0304\ \text{min}^{-1} = 5.07 \times 10^{-4}\ \text{s}^{-1}}\]