Question

Calculate the time taken in seconds for $40\%$ completion of a first order reaction, if its rate constant is $3.3\times {{10}^{-4}}^{}{{\sec }^{-1}}^{}.$

Answer 1

The above question can be solved as:

Step 1:

Rate constant $=3.3\times {10}^{-4}{\sec }^{-1}$

Since the unit of rate constant is ${\sec }^{-1}$, Hence it is a first order reaction.

Step 2:

For first order reaction:

$\mathrm{k}=\frac{1}{t}\ln \frac{\left[A_0\right]}{\left[A\right]}$ (where k= rate constant, t=time taken, $\left[{\mathrm{A}}_0\right]$ is initial concentration and $\left[\mathrm{A}\right]$ is the concentration at time t.)

Step 3:

To find the time taken for a first order reaction:

$$\begin{gathered} t=\frac{1}{k}\ln \frac{\left[A_0\right]}{\left[A\right]} \\ t=\frac{2.303}{k}\log \frac{\left[A_0\right]}{\left[A\right]} \\ =\frac{1\times 2.303}{3.33\times {10}^{-4}}\log \frac{\left[5\right]}{\left[3\right]} \\ =\frac{2.303}{3.33\times {10}^{-4}}\times 0.22 \\ =30\times {10}^3\times 0.506 \\ t=1518sec. \end{gathered}$$

Therefore, time taken for the above reaction is $1518$ second.