JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
For a reaction scheme. \(A\xrightarrow{{{k_1}}}B\xrightarrow{{{k_2}}}C\) if the rate of formation of \(B\) is set to be zero then the concentration of \(B\) is given by
- A \(\left( {\frac{{{k_1}}}{{{k_2}}}} \right)\left[ A \right]\)
- B \(\left( {{k_1} - {k_2}} \right)\left[ A \right]\)
- C \({k_1}{k_2}\left[ A \right]\)
- D \(\left( {{k_1} + {k_2}} \right)\left[ A \right]\)
Answer & Solution
Correct Answer
(A) \(\left( {\frac{{{k_1}}}{{{k_2}}}} \right)\left[ A \right]\)
Step-by-step Solution
Detailed explanation
Applying steady state of approximation \(\frac{d}{{dt}}[B]\, = \,{K_1}[A]\, - \,{K_2}[B]\) \(O\, = \,{K_1}[A]\, - \,{K_2}[B]\) \(\frac{{{K_1}}}{{{K_2}}}[A]\, = \,[B]\)
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