Question

The half-life of a first order chemical reaction is \(60\) hrs at \(300\,\text{K}\). As temperature is increased to \(310\,\text{K}\), half-life becomes \(40\) hrs. Determine the half-life of the same reaction at \(350\,\text{K}\).

• A. \(10\) min
• B. \(160\) min
• C. \(600\) min
• D. \(6\) hrs

Hint:

For first order reactions: \[ t_{1/2} \propto \frac{1}{k} \] An increase in temperature increases the rate constant exponentially, leading to a sharp decrease in half-life.

Answer 1

The Correct Option is C

Step 1: For a first order reaction, \[ t_{1/2} = \frac{0.693}{k} \] Hence, \[ t_{1/2} \propto \frac{1}{k} \]
Step 2: From the given data at \(300\,\text{K}\) and \(310\,\text{K}\): \[ \frac{k_{310}}{k_{300}} = \frac{t_{300}}{t_{310}} = \frac{60}{40} = 1.5 \]
Step 3: Using Arrhenius equation: \[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \] \[ \ln(1.5) = \frac{E_a}{R}\left(\frac{1}{300} - \frac{1}{310}\right) \] \[ E_a \approx 31.3\,\text{kJ mol}^{-1} \]
Step 4: Now calculate the rate constant ratio between \(350\,\text{K}\) and \(300\,\text{K}\): \[ \ln\left(\frac{k_{350}}{k_{300}}\right) = \frac{E_a}{R}\left(\frac{1}{300} - \frac{1}{350}\right) \] \[ \frac{k_{350}}{k_{300}} \approx e^{1.79} \approx 6 \]
Step 5: Since half-life is inversely proportional to rate constant: \[ t_{350} = \frac{t_{300}}{6} = \frac{60}{6} = 10\,\text{hrs} \] \[ 10\,\text{hrs} = 600\,\text{min} \]