Question

Rate of formation of ${SO}_3$ according to the reaction $2{SO}_2+O_2\to 2{SO}_{3}is1.6\times {10}^{-3}kgmi{n}^{-1}$. Hence rate at which $S{O}_2$ reacts is:

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

Answer 1

Answer: (D)

$1.28\times {10}^{-3}kg{min}^{-1}$

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

$\frac{1}{2}\frac{d\left[S{O}_3\right]}{dt}=1.6\times {10}^{-3}kg/min=\frac{1.6\times {10}^{-3}}{80}mol/min$ $[\because M_{S{O}_3}=80\times {10}^{-3}Kg/mol]$

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

$=-\frac{1.6\times {10}^{-3}}{80\times 10}$ mol/min

$=-\frac{1}{50}\times 64\times {10}^{-3}=-1.28\times {10}^{-3}$ kg/min