Question

A certain first-order reaction is \(45.0 \%\) complete in \(65 \mathrm{~s}\). What are the values of the rate constant and the half-life for this process?

Short answer

The rate constant for this first-order reaction is approximately \(k \approx 0.01169 s^{-1}\). The half-life for this first-order reaction is approximately \(t_{1/2} \approx 59.27s\).

Step-by-step solution

Determine the reaction completion and time elapsed

We are given that the reaction is 45.0% complete in 65 seconds.

Use the formula for first-order reaction

We can use the first-order reaction formula: \[ln(\frac{A_0}{A_t}) = kt\] Where: - \(A_0\) is the initial concentration - \(A_t\) is the concentration at time t - \(k\) is the rate constant - \(t\) is the time elapsed Since the reaction is 45.0% complete, we have 55.0% remaining. Assuming the initial concentration is 1, we can say that \(A_t = 0.55\). Now, substitute the known values into the formula: \[ln(\frac{1}{0.55}) = k \cdot 65s\]

Solve for the rate constant, k

We rearrange the formula to solve for k: \[k = \frac{ln(\frac{1}{0.55})}{65s}\] Now calculate k: \[k \approx 0.01169 s^{-1}\] So the rate constant for this first-order reaction is approximately 0.01169 s^(-1).

Determine the half-life (t) for a first-order reaction

The half-life formula for a first-order reaction can be written as: \[t_{1/2} = \frac{ln(2)}{k}\] Now, substitute the calculated value of k into the formula: \[t_{1/2} = \frac{ln(2)}{0.01169s^{-1}}\]

Calculate the half-life

Finally, calculate the half-life: \[t_{1/2} \approx 59.27s\] So the half-life for this first-order reaction is approximately 59.27 seconds.