If the rate constant of a reaction is 0.03 s^–1, how much time does it take for 7.2 mol L^–1 concentration of the reactant to get reduced to 0.9 mol L–1?

(Given: log 2 = 0.301)

The correct answer is 69.3 s.

Step 1: Identify the order of the reaction
The unit of the rate constant is given as s^-1. Since the unit of the rate constant for a first-order reaction is time^-1, this reaction is a first-order reaction.

Step 2: Recall the integrated rate equation for a first-order reaction
The formula relating the time, rate constant, initial concentration, and final concentration for a first-order reaction is:

$t=\frac{2.303}{k}\log \left(\frac{{\left[R\right]}_0}{\left[R\right]}\right)$

Where:
- $t$ is the time taken,
- $k$ is the rate constant,
- ${\left[R\right]}_0$ is the initial concentration of the reactant,
- $\left[R\right]$ is the concentration of the reactant at time $t$.

Step 3: Substitute the given values into the equation
From the question, we are given:
- Rate constant:
$k=0.03{\text{s}}^{-1}$
- Initial concentration of reactant:
${\left[R\right]}_0=7.2{\text{mol L}}^{-1}$
- Final concentration of reactant:
$\left[R\right]=0.9{\text{mol L}}^{-1}$
- Logarithm base 10 value:
$\log 2=0.301$

Substituting these values into the first-order rate equation, we get:

$t=\frac{2.303}{0.03}\log \left(\frac{7.2}{0.9}\right)$

Step 4: Perform the calculations
First, simplify the ratio of the concentrations:

$\frac{7.2}{0.9}=8$

So, the expression becomes:

$t=\frac{2.303}{0.03}\log 8$

We can express 8 as 2³. Using the properties of logarithms, we have:

$\log 8=\log \left(2^3\right)=3\log 2$

Substitute the given value of log 2 = 0.301:

$\log 8=3\times 0.301=0.903$

Now, substitute this value back into the calculation for time (t):

$t=\frac{2.303\times 0.903}{0.03}$

$t=\frac{2.0796}{0.03}$

$t=69.32\text{s}$

Rounding to three significant figures gives:

$t\approx 69.3\text{s}$

Thus, the time required for the concentration of the reactant to reduce to 0.9 mol L^-1 is 69.3 s.