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Chemistry Question on Chemical Kinetics

The reaction,2SO2(g)+O2(g)⇋2SO3(g)2SO_{2(g)} + O_{2(g)} \leftrightharpoons 2SO_{3(g)} is carried out in a 1 dm31\, dm^3 vessel and 2 dm32 \,dm^3 vessel separately. The ratio of the reaction velocities will be :

A

4:01

B

8:01

C

1:08

D

1:04

Answer

8:01

Explanation

Solution

2SO2( g)+O2( g)⇌2SO3( g)2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})

For this reaction, rate

r1=k[SO2]2[O2]\mathrm{r}_{1}=\mathrm{k}\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]

On doubling the volume of vessel, concentration would be half.

Hence,
Rate(r2)=k([SO2]2)2([O2]2)=r18\left(\mathrm{r}_{2}\right)=\mathrm{k}\left(\frac{\left[\mathrm{SO}_{2}\right]}{2}\right)^{2}\left(\frac{\left[\mathrm{O}_{2}\right]}{2}\right)=\frac{\mathrm{r}_{1}}{8}
r1r2=8:1\frac{r_{1}}{r_{2}}=8: 1