A first order reaction takes 40 min for `30%` decomposition. Calculate `t_(1//2)`. (Given `log 7 =0.845)`

Options

A. 77.7 min
B. 52.5 min
C. 46.2min
D. 22.7 min

Explanation

To solve the problem, we need to calculate the half-life (t_(1/2)) of a first-order reaction given that it takes 40 minutes for 30% decomposition. We will follow these steps:

### Step 1: Understand the given information
- Time (t) = 40 minutes
- Percentage decomposition = 30%
- Initial concentration (A) = A (we will assume this as a variable)
- Amount decomposed (X) = 30% of A = 0.30A
- Remaining concentration (A - X) = A - 0.30A = 0.70A

### Step 2: Use the first-order reaction rate equation
For a first-order reaction, the rate constant (k) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{A}{A - X} \right) \]

Substituting the values we have:
- \( A = A \)
- \( A - X = 0.70A \)
- \( t = 40 \) minutes

Thus, we can rewrite the equation as:
\[ k = \frac{2.303}{40} \log \left( \frac{A}{0.70A} \right) \]

### Step 3: Simplify the logarithm
The logarithm simplifies as follows:
\[ \log \left( \frac{A}{0.70A} \right) = \log \left( \frac{1}{0.70} \right) = \log(1.42857) \]

Using the change of base property, we can express this as:
\[ \log(1.42857) = \log(7) - \log(5) \]
However, for simplicity, we can directly use:
\[ \log(1.42857) \approx 0.155 \] (or we can use the given log values)

### Step 4: Calculate k
Now substituting the value back into the equation:
\[ k = \frac{2.303}{40} \cdot 0.155 \]
Calculating this gives:
\[ k \approx 0.00892 \, \text{min}^{-1} \]

### Step 5: Calculate the half-life (t_(1/2))
For a first-order reaction, the half-life is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]

Substituting the value of k:
\[ t_{1/2} = \frac{0.693}{0.00892} \]

Calculating this gives:
\[ t_{1/2} \approx 77.7 \, \text{minutes} \]

### Final Answer
Thus, the half-life of the reaction is approximately **77.7 minutes**.

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