Read the passage carefully and answer the questions.
The chemical reactions proceed at different rates which depend on various factors like concentration, temperature, etc. Rate law is an expression which relates the rate of reaction with concentration of various reacting species in the reaction. For a first order reaction, the concentration of the reactant after time \( t \) is related with its initial concentration by the relation, \( C = C_0e^{-kt} \). The rate constants for different orders have different expressions. For a first order reaction, \( k =\frac{2.303}{t}\log\frac{a}{a-x} \), where \( a \) is the initial concentration and \( x \) is the extent of reaction.
A reaction which is first order with respect to the reactant A has a rate constant of \( 6\, min^{-1} \). If the reaction is started with \( [A] = 5.0\, mol\, L^{-1} \) when would [A] reach the value of \( 0.05\, mol\, L^{-1} \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.7676 min
For a first order reaction the integrated rate equation gives the time needed for the concentration to fall from \( [\mathrm{A}]_0 \) to \( [\mathrm{A}] \):
\( \displaystyle t = \frac{2.303}{k}\log\frac{[\mathrm{A}]_0}{[\mathrm{A}]} \)
Here \( k = 6\ \mathrm{min^{-1}} \), \( [\mathrm{A}]_0 = 5.0\ \mathrm{mol\,L^{-1}} \) and \( [\mathrm{A}] = 0.05\ \mathrm{mol\,L^{-1}} \).
\( \displaystyle t = \frac{2.303}{6\ \mathrm{min^{-1}}}\log\frac{5.0}{0.05} = \frac{2.303}{6}\log 100 \)
\( \displaystyle t = \frac{2.303 \times 2}{6} = 0.7677\ \mathrm{min} \)
So [A] reaches 0.05 mol L−1 after about 0.7676 min. Because the unit of k is min−1, the time comes out directly in minutes.
Shortcut: For a first order reaction every tenfold fall in concentration takes the same time, \( 2.303/k = 2.303/6 = 0.384 \) min. From 5.0 to 0.05 is a hundredfold fall, that is, two tenfold steps: 2 × 0.384 = 0.768 min.