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-by-step Explanation:
1. Identify the Order of the Reaction:
The unit of the rate constant is given as $s^{-1}$. A rate constant with units of reciprocal time indicates a first-order reaction.

2. First-Order Integrated Rate Equation:
For a first-order reaction, the relationship between time ($t$), the rate constant ($k$), the initial concentration (${\mathrm{[R]}}_0$), and the final concentration at time $t$ ($\mathrm{[R]}$) is given by the integrated rate law:

$t=\frac{2.303}{k}{\log }_{10}\left(\frac{{\mathrm{[R]}}_0}{\mathrm{[R]}}\right)$

3. Substitute the Given Values:
We are given:
- Rate constant, $k=0.03s^{-1}$
- Initial concentration, ${\mathrm{[R]}}_0=7.2{\mathrm{mol\; L}}^{-1}$
- Final concentration, $\mathrm{[R]}=0.9{\mathrm{mol\; L}}^{-1}$
- ${\log }_{10}2=0.301$

Let us calculate the ratio of the concentrations:

$\frac{{\mathrm{[R]}}_0}{\mathrm{[R]}}=\frac{7.2}{0.9}=8$

4. Calculate the Logarithmic Term:
Since $8=2^3$, we can simplify the logarithm as follows:

${\log }_{10}\left(8\right)={\log }_{10}\left(2^3\right)=3{\log }_{10}\left(2\right)$

Using the given value of ${\log }_{10}2=0.301$:

${\log }_{10}\left(8\right)=3\times 0.301=0.903$

5. Calculate the Time ($t$):
Now, substitute these values back into the rate equation:

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

$t=2.303\times 30.1$

$t\approx 69.3\text{s}$

Thus, the time required for the concentration to reduce from 7.2 mol L^-1 to 0.9 mol L^-1 is 69.3 s.