A reaction proceeds through the formation of an intermediate B in a unimolecular reaction \[A\xrightarrow{{{k_a}}}B\xrightarrow{{{k_b}}}C\]
The integrated rate law for this reaction is
A. $$\left[ A \right] = {\left[ A \right]_0}{e^{ - {k_a}t}}$$
B. $$\left[ A \right] = {\left[ A \right]_0}\left( {{e^{ - {k_a}t}} - {e^{ - {k_b}t}}} \right)$$
C. $$\left[ A \right] = \frac{{{{\left[ A \right]}_0}}}{2}\left( {1 + \frac{{{k_a}{e^{ - {k_b}t}} - {k_b}{e^{ - {k_a}t}}}}{{{k_a} - {k_b}}}} \right)$$
D. $$\left[ A \right] = {\left[ A \right]_0}\left( {1 + {e^{ - {k_a}t}} - {e^{ - {k_b}t}}} \right)$$
Answer: Option A
Related Questions on Chemical Kinetics
The half-life of a first order reaction varies with temperature according to
A. In t1/2 ∝ 1/T
B. In t1/2 ∝ T
C. t1/2 ∝ 1/T2
D. t1/2 ∝ T2
A. 0.1386 min-1
B. 0.0693 min-1
C. 0.1386 mol L-1min-1
D. 0.0693 mol L-1min-1
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