Question

$2S{O}_2+O_2\to 2S{O}_3$

Rate of formation of $S{O}_3$ according to the reaction is $1.6\times {10}^{-3}$ kg $mi{n}^{-1}$. Hence rate at which $S{O}_2$ reacts is:

• A. $1.6\times {10}^{-3}$ kg $mi{n}^{-1}$
• B. $8.0\times {10}^{-4}$ kg $mi{n}^{-1}$
• C. $3.2\times {10}^{-3}$ kg $mi{n}^{-1}$
• D. $1.28\times {10}^{-3}$ kg $mi{n}^{-1}$

Answer 1

The correct option is D $1.28\times {10}^{-3}$ kg $mi{n}^{-1}$

${2SO}_2+O_2⟶{2SO}_3.$

According to rate law

$\frac{-1}{2}\frac{d\left[S{O}_2\right]}{dt}=\frac{-d\left[O_2\right]}{dt}=\frac{1}{2}\frac{d\left[{SO}_3\right]}{dt}-\frac{d\left[S{O}_2\right]}{dt}=\frac{d\left[{SO}_3\right]}{dt}$

In unit time, decrease in number of moles of ${SO}_2$ is equivalent to the increase in number of moles of ${SO}_3.$

Rate of from action of $S{O}_3=1.6\times {10}^{-3}kg/min.$

MM of $S{O}_3=80U$

MM of $S{O}_2=64U$

Rate of reaction of $S{O}_2=\frac{64}{80}\times 1.6\times {10}^{-3}kg/min.$

$=1.28\times {10}^{-3}kg/min.$

Hence, the correct option is $D$