Solveeit Logo
Login

Question

Chemistry Question on Rate of a Chemical Reaction

For an elementary reaction 2A+3B→4C+D2A+3B \to 4C+D the rate of appearance of CC at time ?t??t? is 2.8×10−3 mol L−1S−12.8\times10^{-3} \,mol \,L^{-1}S^{-1}. Rate of disappearance of BB at ?tt? will be

A

43(2.8×10−3)molL−1S−1\frac{4}{3}(2.8\times10^{-3}) mol L^{-1}S^{-1}

B

34(2.8×10−3)molL−1S−1\frac{3}{4}(2.8\times10^{-3}) mol L^{-1}S^{-1}

C

2(2.8×10−3)molL−1S−12(2.8\times10^{-3}) mol L^{-1}S^{-1}

D

14(2.8×10−3)molL−1S−1\frac{1}{4}(2.8\times10^{-3}) mol L^{-1}S^{-1}

Answer

34(2.8×10−3)molL−1S−1\frac{3}{4}(2.8\times10^{-3}) mol L^{-1}S^{-1}

Explanation

Solution

The given reaction is,

2A+3B→4C+D2 A+3 B \rightarrow 4 C+D

So, −13d[B]dt=14d[c]dt-\frac{1}{3} \frac{d[B]}{d t}=\frac{1}{4} \frac{d[c]}{d t}
⇒−d[B]dt=34d[C]dt\Rightarrow-\frac{d[B]}{d t}=\frac{3}{4} \frac{d[C]}{d t}
=34(2.8×10−3)mol L−1S−1=\frac{3}{4}\left(2.8 \times 10^{-3}\right) m o l\, L^{-1} S^{-1}