`90%` of a first order reaction was completed in `100 min`. How much time it will take for `80%` completion of a reaction

Options

A. `90 min`
B. `80 min`
C. `70 min`
D. `60 min`

Explanation

To solve the problem of how much time it will take for 80% completion of a first-order reaction given that 90% completion takes 100 minutes, we can follow these steps:

### Step 1: Understand the reaction and the data given
We have a first-order reaction where 90% of the reactant has been converted into product in 100 minutes. This means that 10% of the reactant remains.

### Step 2: Set up the equations
For a first-order reaction, the integrated rate law is given by:

\[ t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \]

Where:
- \( t \) is the time taken for the reaction
- \( k \) is the rate constant
- \( [A_0] \) is the initial concentration of the reactant
- \( [A] \) is the concentration of the reactant at time \( t \)

### Step 3: Calculate the rate constant \( k \)
From the information given, when 90% of the reactant is converted, 10% remains. Thus, we can express this as:

\[ [A] = 0.1 [A_0] \]

Substituting into the integrated rate law for \( t = 100 \) minutes:

\[ 100 = \frac{2.303}{k} \log \left( \frac{[A_0]}{0.1 [A_0]} \right) \]

This simplifies to:

\[ 100 = \frac{2.303}{k} \log(10) \]

Since \( \log(10) = 1 \):

\[ 100 = \frac{2.303}{k} \]

Rearranging gives:

\[ k = \frac{2.303}{100} = 0.02303 \, \text{min}^{-1} \]

### Step 4: Calculate time for 80% completion
For 80% completion, 20% of the reactant remains:

\[ [A] = 0.2 [A_0] \]

Now we substitute this into the integrated rate law:

\[ t_{80\%} = \frac{2.303}{k} \log \left( \frac{[A_0]}{0.2 [A_0]} \right) \]

This simplifies to:

\[ t_{80\%} = \frac{2.303}{k} \log(5) \]

### Step 5: Substitute \( k \) and calculate \( t_{80\%} \)
Now substituting \( k \):

\[ t_{80\%} = \frac{2.303}{0.02303} \log(5) \]

Using \( \log(5) \approx 0.699 \):

\[ t_{80\%} = \frac{2.303}{0.02303} \times 0.699 \]

Calculating this gives:

\[ t_{80\%} \approx 70 \, \text{minutes} \]

### Final Answer
Thus, the time required for 80% completion of the reaction is approximately **70 minutes**.

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