If the rate constant of a reaction is $0.03s^{-1}$, 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: $\log 2=0.301$)

For a first‑order reaction the concentration changes with time according to

$t=\frac{1}{k}\ln \left(\frac{\mathrm{C\_0}}{C}\right)$

where

• k is the rate constant (0.03 s⁻¹),
• C_0 is the initial concentration (7.2 mol L⁻¹),
• C is the final concentration (0.9 mol L⁻¹),
• t is the time required.

First compute the concentration ratio:

$\frac{\mathrm{C\_0}}{C}=\frac{7.2}{0.9}=8$

The natural logarithm can be evaluated using the given common‑log value $\log 2=0.301$. Because

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

and the problem supplies $\log 10\left(2\right)=0.301$, we obtain

$\log 10\left(8\right)=30.301=0.903$

Convert the common log to a natural log using $\ln \left(x\right)=2.303\log 10\left(x\right)$:

$\ln \left(8\right)=2.3030.903\approx 2.08$

Now substitute into the time equation:

$t=\frac{1}{0.03}2.08\approx 69.3 s$

Therefore, it takes approximately **69.3 seconds** for the concentration to drop from 7.2 mol L⁻¹ to 0.9 mol L⁻¹.