Question

The following reaction is exothermic.

Answer 1

Answer : The correct option is, (C) spontaneous only at low temperatures.

Explanation :

According to Gibb's equation:

$\Delta G=\Delta H-T\Delta S$

$\Delta G$ = Gibbs free energy  

$\Delta H$ = enthalpy change

$\Delta S$ = entropy change  

T = temperature in Kelvin

As we know that:

$\Delta G$= +ve, reaction is non spontaneous

$\Delta G$= -ve, reaction is spontaneous

$\Delta G$= 0, reaction is in equilibrium

The given chemical reaction is:

$2S(s)+3O_2(g)\rightarrow 2SO_3(g)$

As we are given that, the given reaction is exothermic that means the enthalpy change is negative.

In this reaction, the randomness of reactant molecules are more and as we move towards the formation of product the randomness become less that means the degree of disorderedness decreases. So, the entropy will also decreases that means the change in entropy is negative.

Now we have to determine the spontaneity of this reaction when ΔH is negative and ΔS is negative.

As, $\Delta G=\Delta H-T\Delta S$

$\Delta G=(-ve)-T(-ve)$

$\Delta G=(+ve)$   (at high temperature) (non-spontaneous)

$\Delta G=(-ve)$   (at low temperature) (spontaneous)

Thus, the reaction is spontaneous only at low temperatures.