Question

For a first order reaction \(A → B\), the rate constant, \(k = 5.5×10^{–14}s^{–1}\). The time required for 67% completion of reaction is \(x×10^{–1}\) times the half life of reaction. The value of \(x\) is _____. (Nearest integer)

Answer 1

Correct Answer: 16

A first-order reaction is characterized by the equation for decay, which relates the concentration of reactants over time. The general formula for the time required for completion is:

\[ t = \frac{\ln(1-f)}{-k} \]

where \( f \) is the fraction of the reaction completed, and \( k \) is the rate constant.

Step 1: Calculate the time for 67% completion

Here, \( f = 0.67 \); thus, the equation becomes:

\[ t_{0.67} = \frac{\ln(1-0.67)}{-5.5\times10^{-14}} \]

Compute the natural logarithm:

\[ \ln(1-0.67) = \ln(0.33) \approx -1.10866 \]

Substitute \(\ln(0.33)\) into the formula:

\[ t_{0.67} = \frac{-1.10866}{-5.5\times10^{-14}} \]

Simplify:

\[ t_{0.67} \approx 2.0157\times10^{13} \]\text{s}

Step 2: Calculate the half-life \( t_{1/2} \)

The half-life formula for a first-order reaction is:

\[ t_{1/2} = \frac{\ln(2)}{k} \]

\[ \ln(2) \approx 0.693 \]

Therefore:

\[ t_{1/2} = \frac{0.693}{5.5\times10^{-14}} \approx 1.26\times10^{13} \]\text{s}

Step 3: Calculate \( x \)

The problem states:

\[ t_{0.67} = x\times10^{-1}\times t_{1/2} \]

Substitute \( t_{0.67} \) and \( t_{1/2} \):

\[ 2.0157\times10^{13} = x\times10^{-1}\times1.26\times10^{13} \]

Solve for \( x \):

\[ x = \frac{2.0157}{1.26} \times 10 \approx 16.00 \]

Validation

The computed value \( x = 16 \) is within the specified range [16,16].

Therefore, the value of \( x \) is 16.