JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
For a reaction \(\mathrm{A} \xrightarrow{\mathrm{K}_4} \mathrm{~B} \xrightarrow{\mathrm{K}_2} \mathrm{C}\) If the rate of formation of \(B\) is set to be zero then the concentration of \(B\) is given by _______.
- A \(\mathrm{K}_1 \mathrm{~K}_2[\mathrm{~A}]\)
- B \(\left(\mathrm{K}_1-\mathrm{K}_2\right)[\mathrm{A}]\)
- C \(\left(\mathrm{K}_1+\mathrm{K}_2\right)[\mathrm{A}]\)
- D \(\left(\mathrm{K}_1 / \mathrm{K}_2\right)[\mathrm{A}]\)
Answer & Solution
Correct Answer
(D) \(\left(\mathrm{K}_1 / \mathrm{K}_2\right)[\mathrm{A}]\)
Step-by-step Solution
Detailed explanation
Rate of formation of \(B\) is \(\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}=\mathrm{k}_1[\mathrm{~A}]-\mathrm{k}_2[\mathrm{~B}]\) \(0=\mathrm{k}_1[\mathrm{~A}]-\mathrm{k}_2[\mathrm{~B}]\) \(\left(\frac{\mathrm{k}_1}{\mathrm{k}_2}\right)[\mathrm{A}]=[\mathrm{B}]\)
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