For a first order reaction, rate constant is given is `log k = 14 - (1.2 xx 10^(4))/(T)`, then what will be value of temperature if
its half life period is `6.93 xx 10^(-3) min` ?

Options

A. `100 K`
B. `1000 K`
C. `720 K`
D. `327 K`

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant equations for a first-order reaction.

### Step 1: Understand the relationship between half-life and rate constant
For a first-order reaction, the half-life (\(T_{1/2}\)) is related to the rate constant (\(k\)) by the formula:
\[ T_{1/2} = \frac{0.693}{k} \]

### Step 2: Calculate the rate constant \(k\)
Given the half-life period is \(T_{1/2} = 6.93 \times 10^{-3} \text{ min}\), we can rearrange the formula to find \(k\):
\[ k = \frac{0.693}{T_{1/2}} \]
Substituting the value of \(T_{1/2}\):
\[ k = \frac{0.693}{6.93 \times 10^{-3}} \text{ min}^{-1} \]

### Step 3: Perform the calculation for \(k\)
Calculating \(k\):
\[ k = \frac{0.693}{6.93 \times 10^{-3}} \approx 100 \text{ min}^{-1} \]

### Step 4: Use the given equation for \(k\)
We have the equation for the rate constant:
\[ \log k = 14 - \frac{1.2 \times 10^4}{T} \]
Substituting \(k = 100\):
\[ \log 100 = 14 - \frac{1.2 \times 10^4}{T} \]

### Step 5: Calculate \(\log 100\)
\[ \log 100 = 2 \quad (\text{since } 100 = 10^2) \]
Now substituting this into the equation:
\[ 2 = 14 - \frac{1.2 \times 10^4}{T} \]

### Step 6: Rearrange the equation to solve for \(T\)
Rearranging gives:
\[ \frac{1.2 \times 10^4}{T} = 14 - 2 \]
\[ \frac{1.2 \times 10^4}{T} = 12 \]

### Step 7: Solve for \(T\)
Now, multiplying both sides by \(T\) and rearranging:
\[ 1.2 \times 10^4 = 12T \]
\[ T = \frac{1.2 \times 10^4}{12} \]
Calculating \(T\):
\[ T = 1000 \text{ K} \]

### Final Answer
The temperature \(T\) is \(1000 \text{ K}\).

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