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. Lowering of temperature as well as pressure
• D. Lowering the pressure and increasing the temperature

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 on 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.