Question

For the reaction involving the formation of sulphur trioxide from sulphur dioxide and oxygen, the rate of reaction with respect to reactants is given as:

• A. $r=\frac{1}{2}\frac{\Delta \left[O_2\right]}{\Delta t}$
• B. $r=\frac{\Delta \left[{SO}_2\right]}{\Delta t}$
• C. $r=\frac{-\Delta \left[{SO}_2\right]}{\Delta t}$
• D. $r=-\frac{1}{2}\frac{\Delta \left[{SO}_2\right]}{\Delta t}$

Answer 1

Answer: (D)

$r=-\frac{1}{2}\frac{\Delta \left[{SO}_2\right]}{\Delta t}$

Reaction is $2S{O}_2+O_2\rightleftharpoons 2S{O}_3$
In the reaction an excess of oxygen relative to the proportions is demanded by the equation. According to Le Chatelier's Principle, Increasing the concentration of oxygen in the mixture causes the position of equilibrium to shift towards the right.
Thus concentration of oxygen does not affect the rate of reaction. $2moles$ of $S{O}_2$ are involved in the formation of $S{O}_3$ and hence rate of reaction is
$r=-\frac{1}{2}\frac{\Delta \left[{SO}_2\right]}{\Delta t}$