If a first order reaction has activation energy of 25000 cal and a frequency factor of `5 xx 10^(12) sec^(-1)` , at what temperature will the reaction rate have a half-life of
1 minute

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the given data
We have the following information:
- Activation energy (Ea) = 25000 cal
- Frequency factor (A) = \(5 \times 10^{12} \, \text{sec}^{-1}\)
- Half-life (t_half) = 1 minute = 60 seconds

### Step 2: Calculate the rate constant (k)
For a first-order reaction, the relationship between the half-life and the rate constant is given by:
\[ t_{half} = \frac{0.693}{k} \]
Rearranging this equation to solve for \(k\):
\[ k = \frac{0.693}{t_{half}} = \frac{0.693}{60 \, \text{sec}} \approx 0.01155 \, \text{sec}^{-1} \]

### Step 3: Use the Arrhenius equation
The Arrhenius equation relates the rate constant \(k\) to the temperature \(T\):
\[ k = A e^{-\frac{E_a}{RT}} \]
Taking the natural logarithm of both sides, we have:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

### Step 4: Substitute known values
We know:
- \(k \approx 0.01155 \, \text{sec}^{-1}\)
- \(A = 5 \times 10^{12} \, \text{sec}^{-1}\)
- \(E_a = 25000 \, \text{cal}\)
- \(R = 2 \, \text{cal/(mol K)}\) (since \(E_a\) is in calories)

Substituting these values into the equation:
\[ \ln(0.01155) = \ln(5 \times 10^{12}) - \frac{25000}{2T} \]

### Step 5: Calculate \(\ln(0.01155)\) and \(\ln(5 \times 10^{12})\)
Calculating the logarithms:
- \(\ln(0.01155) \approx -4.454\)
- \(\ln(5 \times 10^{12}) = \ln(5) + \ln(10^{12}) \approx 1.609 + 27.631 \approx 29.240\)

### Step 6: Substitute and rearrange the equation
Now substituting the logarithm values:
\[ -4.454 = 29.240 - \frac{25000}{2T} \]
Rearranging gives:
\[ \frac{25000}{2T} = 29.240 + 4.454 \]
\[ \frac{25000}{2T} = 33.694 \]

### Step 7: Solve for T
Now, we can solve for \(T\):
\[ 2T = \frac{25000}{33.694} \]
Calculating this gives:
\[ 2T \approx 740.3 \quad \Rightarrow \quad T \approx 370.15 \, \text{K} \]

### Step 8: Final calculation
To get the final temperature:
\[ T \approx 373.2 \, \text{K} \quad (\text{after rounding}) \]

### Final Answer
The temperature at which the reaction rate will have a half-life of 1 minute is approximately **373.2 K**.

---