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Chemistry Question on Chemical Kinetics

In a reaction 2A+B⟶3C2 A+B \longrightarrow 3 C, the concentration of AA decreases from 0.5 mol L −10.5\, mol \,L \,^{-1} to 0.3 mol L−10.3\, mol\, L^{-1} in 1010 minutes. The rate of production of CC during this period is

A

0.01 mol L−1 min−10.01\, mol \,L^{-1}\,min^{-1}

B

0.04 mol L−1 min−10.04 \,mol\, L^{-1}\, min^{-1}

C

0.05 mol L−1 min−10.05 \,mol \,L^{-1}\, min^{-1}

D

0.03 mol L−1 min−10.03\, mol\, L^{-1}\, min^{-1}

Answer

0.03 mol L−1 min−10.03\, mol\, L^{-1}\, min^{-1}

Explanation

Solution

The correct option is(D): 0.03 mol L−1 min−10.03\, mol\, L^{-1}\, min^{-1}.

Rate of decrease in concentration of AA
−(dAdt)=0.5−0.310=0.210=0.02-\left(\frac{d A}{d t}\right)=\frac{0.5-0.3}{10}=\frac{0.2}{10}=0.02

From the equation

2A+B⟶3 C2 A+B \longrightarrow 3 \,C
−d[A]2×dt=−d[B]dt=+d[C]3 dt-\frac{d[A]}{2 \times d t}=-\frac{d[B]}{d t}=+\frac{d[C]}{3 \,d t}

Thus, the rate of production of CC,

d[c]dt=32×d[c]dt\frac{d[c]}{d t}=\frac{3}{2} \times \frac{d[c]}{d t}
=32×=\frac{3}{2} \times rate of decrease in concentration of AA
=32×0.02=0.03 mol L−1 min⁡−1=\frac{3}{2} \times 0.02=0.03\, mol\, L ^{-1} \,\min ^{-1}