Question

Consider the reaction : $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right);△H^o=-198kJ$
The favourable conditions for the forward reaction are:

• A. Increasing temperature as well as pressure
• B. Lowering the temperature and increasing the pressure
• C. Any value of temperature and pressure
• D. Lowering of temperature as well as pressure

Answer 1

The correct option is B Lowering the temperature and increasing the pressure

As per the given reaction,
$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right);△H^o=-198kJ$

Number of moles of reactants are more than the number of moles of products.
So, on increasing the pressure the reaction as per Le - Chatelier principle, will tend to shift in that direction where the number of moles are decreasing . Here number of moles are decreasing in the forward direction.So increasing pressure is favourable for forward reaction.

Effect of temperature:
We know Van't Hoff equation,
$\frac{dlnK}{dT}=\frac{\Delta H^o}{R{T}^2}$
on integrating we get:
$ln\left(\frac{K_1}{K_2}\right)=\frac{\Delta H^o}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

$\Delta H^o$ is negative here (exothermic)
Since, $(△H^o=-198kJ$)

Case 1:
When temperature $\uparrow \Rightarrow T_2>T_1$
$\Rightarrow$ RHS of the equation is positive.
$ln\left(\frac{K_1}{K_2}\right)>0\Rightarrow K_1>K_2$
$\Rightarrow$Product amount decreases on increasing temperature.
$\Rightarrow$Backward reaction is preffered on increasing temperature.


Case 2:
When temperature $\downarrow \Rightarrow T_2<T_1$
$\Rightarrow$ RHS of the equation is negative.
$ln\left(\frac{K_1}{K_2}\right)<0\Rightarrow K_1<K_2$
$\Rightarrow$Product amount increases on decreasing temperature.
$\Rightarrow$Forward reaction is preffered on decreasing temperature.