Question

If the rate constant of a reaction is 0.03 s$^{-1}$, how much time does it take for a 7.2 mol L$^{-1}$ concentration of the reactant to get reduced to 0.9 mol L$^{-1}$?
(Given: log 2 = 0.301)

• A. 23.1 s
• B. 210 s
• C. 21.0 s
• D. 69.3 s

Hint:

For a first-order reaction, the relationship between concentration and time is logarithmic. Always ensure to use the integrated rate law for calculations involving concentration changes over time.

Answer 1

The Correct Option is D

The units of the rate constant are s\textsuperscript{-1}, indicating a first-order reaction. The integrated rate law for a first-order reaction is: $$ \ln{\frac{[A]_t}{[A]_0}} = -kt $$ Where: $[A]_t$ is the concentration of reactant at time t. It is given as 0.9 mol L\textsuperscript{-1}. $[A]_0$ is the initial concentration of reactant. It is given as 7.2 mol L\textsuperscript{-1}. k is the rate constant, which is 0.03 s\textsuperscript{-1}. t is the time. Substituting the given values into the integrated rate law: $$ \ln{\frac{0.9}{7.2}} = -0.03t $$ $$ \ln{\frac{1}{8}} = -0.03t $$ $$ \ln{1} - \ln{8} = -0.03t $$ $$ 0 - \ln{2^3} = -0.03t $$ $$ -3\ln{2} = -0.03t $$ $$ t = \frac{3\ln{2}}{0.03} $$ Given that log 2 = 0.301, the natural logarithm of 2 can be calculated: $$ \ln{2} = 2.303 \times \log{2} = 2.303 \times 0.301 \approx 0.693 $$ Substituting the value of $\ln{2}$: $$ t = \frac{3 \times 0.693}{0.03} = \frac{2.079}{0.03} = 69.3 \text{ s} $$ The time required is approximately 69.3 s. The correct answer is (4).