KCET · Chemistry · p Block Elements (Group 15, 16, 17 & 18)
For a reaction \( \frac{1}{2} \mathrm{~A} \rightarrow 2 \mathrm{~B} \) rate of disappearance of \( \mathrm{A} \) is related to rate of appearance of \( \mathrm{B} \) by the
expression
- A \( \frac{-d[A]}{d t}=4 \frac{d[B]}{d t} \)
- B \( \frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{d} \mathrm{t}}=\frac{1}{4} \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}} \)
- C \( \frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{d} \mathrm{t}}=\frac{1}{2} \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}} \)
- D \( \frac{-d[A]}{d t}=\frac{d[B]}{d t} \)
Answer & Solution
Correct Answer
(B) \( \frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{d} \mathrm{t}}=\frac{1}{4} \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}} \)
Step-by-step Solution
Detailed explanation
For the given reaction,
\( \frac{1}{2} A \rightarrow 2 B \) or \( A \rightarrow 4 B \)
Rate of disappearance of \( A=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}} \)
Rate of appearance of \( B=-\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}} \)
Rate of disappearance of \( A= \) Rate of appearance of \( B \).
Therefore,
\( -\frac{d[A]}{d t}=\frac{1}{4} \frac{d[B]}{d t} \)
\( \frac{1}{2} A \rightarrow 2 B \) or \( A \rightarrow 4 B \)
Rate of disappearance of \( A=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}} \)
Rate of appearance of \( B=-\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}} \)
Rate of disappearance of \( A= \) Rate of appearance of \( B \).
Therefore,
\( -\frac{d[A]}{d t}=\frac{1}{4} \frac{d[B]}{d t} \)
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