Question

For an elementary reaction $2A+3B \to 4C+D$ the rate of appearance of $C$ at time $t$ is $2.8\times10^{-3} \,mol \,L^{-1}S^{-1}$. Rate of disappearance of $B$ at $t$ will be

• A. $\frac{4}{3}(2.8\times10^{-3}) mol L^{-1}S^{-1}$
• B. $\frac{3}{4}(2.8\times10^{-3}) mol L^{-1}S^{-1}$
• C. $2(2.8\times10^{-3}) mol L^{-1}S^{-1}$
• D. $\frac{1}{4}(2.8\times10^{-3}) mol L^{-1}S^{-1}$

Answer 1

The Correct Option is B

The given reaction is, 

\(2 A+3 B \rightarrow 4 C+D\) 

So, \(-\frac{1}{3} \frac{d[B]}{d t}=\frac{1}{4} \frac{d[c]}{d t}\)

\(\Rightarrow-\frac{d[B]}{d t}=\frac{3}{4} \frac{d[C]}{d t}\)

\(=\frac{3}{4}\left(2.8 \times 10^{-3}\right) m o l\, L^{-1} S^{-1}\)

Answer 2

The given reaction is: \[ 2A + 3B \rightarrow 4C + D \] For elementary reactions, the rate of appearance or disappearance is directly proportional to the stoichiometric coefficients. Since 3 moles of B react for every 4 moles of C, the rate of disappearance of B is related to the rate of appearance of C by the ratio of their stoichiometric coefficients: \[ \text{Rate of disappearance of B} = \frac{3}{4} \times \text{Rate of appearance of C} \] Given that the rate of appearance of C is \( 2.8 \times 10^{-3} \) mol L\(^{-1}\) s\(^{-1}\), we can calculate the rate of disappearance of B as: \[ \text{Rate of disappearance of B} = \frac{3}{4} \times (2.8 \times 10^{-3}) = \frac{4}{3} \times (2.8 \times 10^{-3}) \, \text{mol L}^{-1} \, \text{s}^{-1} \] Thus, the rate of disappearance of B at time ‘t’ is \( \frac{4}{3} \times (2.8 \times 10^{-3}) \) mol L\(^{-1}\) s\(^{-1}\).