For a first order reaction 75% reaction complete in 90 min. for the same reaction 60% complete in what time?

Options

A. 45 min
B. 37 min
C. 60 min
D. 40 min

Answer

To solve the problem, we will use the first-order reaction kinetics formula. Let's break down the steps:

### Step 1: Understand the first-order reaction formula
The formula for the time taken for a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{A_0}{A_t} \right) \]
where:
- \( A_0 \) is the initial concentration,
- \( A_t \) is the concentration at time \( t \),
- \( k \) is the rate constant.

### Step 2: Calculate for 75% completion
For the first scenario where 75% of the reaction is complete in 90 minutes:
- \( A_0 = 100\% \)
- \( A_t = 100\% - 75\% = 25\% \)

Using the formula:
\[ t_{75\%} = \frac{2.303}{k} \log \left( \frac{100}{25} \right) \]
\[ t_{75\%} = \frac{2.303}{k} \log(4) \]

### Step 3: Calculate for 60% completion
For the second scenario where we need to find the time for 60% completion:
- \( A_0 = 100\% \)
- \( A_t = 100\% - 60\% = 40\% \)

Using the formula:
\[ t_{60\%} = \frac{2.303}{k} \log \left( \frac{100}{40} \right) \]
\[ t_{60\%} = \frac{2.303}{k} \log(2.5) \]

### Step 4: Set up the ratio of times
Since both equations have the same \( \frac{2.303}{k} \) term, we can set up a ratio:
\[ \frac{t_{60\%}}{t_{75\%}} = \frac{\log(2.5)}{\log(4)} \]

### Step 5: Substitute known values
We know \( t_{75\%} = 90 \) minutes, so:
\[ t_{60\%} = 90 \times \frac{\log(2.5)}{\log(4)} \]

### Step 6: Calculate the logarithmic values
Using logarithmic values:
- \( \log(4) \approx 0.602 \)
- \( \log(2.5) \approx 0.3979 \)

### Step 7: Calculate \( t_{60\%} \)
Substituting the values:
\[ t_{60\%} = 90 \times \frac{0.3979}{0.602} \]
\[ t_{60\%} \approx 90 \times 0.660 \]
\[ t_{60\%} \approx 59.48 \text{ minutes} \]

### Step 8: Round the answer
Rounding \( 59.48 \) gives us approximately \( 60 \) minutes.

### Final Answer
The time required for 60% completion of the reaction is approximately **60 minutes**.

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