Answer to Question #93411 in Physical Chemistry for Atul

Question

Rate of formation of So3 according to the reaction 2So2 + O2 gives 2So3 is 1.6×10^-3 kg/min hence rate at which So2 reacts as.

Answer 1

$2 SO_2+O_2 \to 2 SO_3$

By rate law kinetics:

$-\frac{1}{2} \frac{d[SO_2]}{dt}=\frac{1}{2} \frac{d[SO_3]}{dt}$

hence,$- \frac{d[SO_2]}{dt}$ $=$ $\frac{d[SO_3]}{dt}$

so,rate at which $SO_2$ reacts=$-($ rate at which $SO_3$ forms.$)$

$=-1.6 \times 10^{-3} \space kg /min$ (negative sign represents the diminishing concentration of $SO_2$ )