The rate constant of a first order reaction of H2O2 is given by log k=14.341.25*10^4k/t`. Calculate Ea for this reaction and at what temperature, the half-life of the reaction is 256 min?

Answer

To solve the problem, we need to calculate the activation energy (Ea) for the reaction and determine the temperature at which the half-life of the reaction is 256 minutes. We'll follow these steps:

### Step 1: Determine the Rate Constant (k) from Half-Life
For a first-order reaction, the half-life (Tf) is given by the formula:
\[ Tf = \frac{0.693}{k} \]
We are given that \( Tf = 256 \) minutes. First, we need to convert this into seconds:
\[ Tf = 256 \, \text{min} \times 60 \, \text{s/min} = 15360 \, \text{s} \]
Now, we can rearrange the half-life formula to find \( k \):
\[ k = \frac{0.693}{Tf} = \frac{0.693}{15360 \, \text{s}} \approx 4.51 \times 10^{-5} \, \text{s}^{-1} \]

### Step 2: Use the Given Logarithmic Equation to Find Activation Energy (Ea)
The rate constant \( k \) is given by the equation:
\[ \log k = 14.34 - \frac{1.25 \times 10^4}{T} \]
We can substitute the value of \( k \) we just calculated into this equation:
\[ \log(4.51 \times 10^{-5}) = 14.34 - \frac{1.25 \times 10^4}{T} \]
Calculating the logarithm:
\[ \log(4.51 \times 10^{-5}) \approx -4.35 \]
Now we can set up the equation:
\[ -4.35 = 14.34 - \frac{1.25 \times 10^4}{T} \]
Rearranging gives:
\[ \frac{1.25 \times 10^4}{T} = 14.34 + 4.35 \]
\[ \frac{1.25 \times 10^4}{T} = 18.69 \]
Now, solving for \( T \):
\[ T = \frac{1.25 \times 10^4}{18.69} \approx 669 \, \text{K} \]

### Step 3: Calculate Activation Energy (Ea)
From the Arrhenius equation:
\[ \log k = \log A - \frac{Ea}{2.303RT} \]
We can rearrange to find \( Ea \):
\[ Ea = -2.303R \cdot (14.34 - \log k) \]
Substituting \( R = 8.314 \, \text{J/mol·K} \):
\[ Ea = -2.303 \times 8.314 \times (14.34 + 4.35) \]
Calculating:
\[ Ea = -2.303 \times 8.314 \times 18.69 \approx 239339 \, \text{J/mol} \approx 239.34 \, \text{kJ/mol} \]

### Final Answers
- Activation Energy, \( Ea \approx 239.34 \, \text{kJ/mol} \)
- Temperature at which half-life is 256 minutes, \( T \approx 669 \, \text{K} \)