KCET · Chemistry · Chemical Kinetics
For an elementary reaction \(2 A+3 B \rightarrow 4 C+D\) the rate of appearance of \(C\) at time ' \(t\) ' is \(2.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~S}^{-1}\). Rate of disappearance of \(B\) at ' \(t\) '. \(t\) will be
- A \(\frac{4}{3}\left(2.8 \times 10^{-3}\right) \mathrm{mol} \mathrm{L}-1 \mathrm{~S}^{-1}\)
- B \(\frac{3}{4}\left(2.8 \times 10^{-3}\right) \mathrm{mol} \mathrm{L}^{-1} \mathrm{~S}^{-1}\)
- C \(2\left(2.8 \times 10^{-3}\right) \mathrm{mol} \mathrm{L}^{-1} \mathrm{~S}^{-1}\)
- D \(\frac{1}{4}\left(2.8 \times 10^{-3}\right) \mathrm{mol} \mathrm{L}-1 \mathrm{~S}^{-1}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{4}\left(2.8 \times 10^{-3}\right) \mathrm{mol} \mathrm{L}^{-1} \mathrm{~S}^{-1}\)
Step-by-step Solution
Detailed explanation
\(2 A+3 B \longrightarrow 4 C+D\)
\(\begin{aligned}
\frac{d(C)}{d t} &=2.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}(\text { at time } t) \\
-\frac{d(B)}{d t} &=? \\
-\frac{1}{3} \frac{d(B)}{\Delta t} &=+\frac{1}{4} \frac{d[C]}{\Delta t}
\end{aligned}\)
Rate of disappearance of \(B=\frac{3}{4} \times\) rate of appearance of \(C\)
\(=\frac{3}{4} \times 2.8 \times 10^{-3}\)
\(\begin{aligned}
\frac{d(C)}{d t} &=2.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}(\text { at time } t) \\
-\frac{d(B)}{d t} &=? \\
-\frac{1}{3} \frac{d(B)}{\Delta t} &=+\frac{1}{4} \frac{d[C]}{\Delta t}
\end{aligned}\)
Rate of disappearance of \(B=\frac{3}{4} \times\) rate of appearance of \(C\)
\(=\frac{3}{4} \times 2.8 \times 10^{-3}\)
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