Q20.6P - Chemistry: Molecular Nature Of Matter And Change

Question

How does the entropy of the surroundings change during an exothermic reaction? An endothermic reaction? Other than the examples in text, describe a spontaneous endothermic process.

Step-by-Step Solution

Answer

In the exothermic process, the entropy of surroundings increases$\to ∆{\text{S}}_{\text{surr}}\text{ > 0}$

In the endothermic process, the entropy of surroundings decreases$\to ∆{\mathrm{S}}_{\mathrm{surr}}<0$

The example of endothermic spontaneous reaction: vinegar ${\text{(CH}}_{\text{3}}\text{COOH)}$ with the baking soda ${\text{(NaHCO}}_{\text{3}}\text{)}$

Step 1: Concept Introduction

An endothermic reaction is one in which heat is required for the reaction to take place. Consider the melting of ice cubes as an example.

An exothermic reaction is a reaction in which heat is released. Condensation of rain from water vapour is an example.

Step 2: Entropy of change during exothermic and endothermic reaction

The exothermic process by definition is a process, characterized by the system heat release to the surroundings.

$∆{\text{S}}_{\text{sys}}\text{ = }\frac{{\text{q}}_{\text{rev}}}{\text{T}}$

Thus, as the temperature of the surroundings increases, the entropy increases

$\to ∆{\text{S}}_{\text{surr}}\text{ > 0}$

 The endothermic process, by definition, is a process, characterized by the system heat absorption from the surroundings. As a result, the temperature of the surroundings decreases, as well as the entropy$\to ∆{\mathrm{S}}_{\mathrm{surr}}<0$

Step 3: Spontaneous endothermic reaction

The spontaneous endothermic process example, that we face frequently is the reaction of vinegar ${\text{(CH}}_{\text{3}}\text{COOH)}$ with the baking soda ${\text{(NaHCO}}_{\text{3}}\text{)}$- even though the majority of acid-base neutralization reactions are exothermic, this is not one of them, it is indeed endothermic.
${\text{NaHCO}}_{\text{3}}\text{(s)} \text{ + } {\text{CH}}_{\text{3}}\text{COOH(aq)}\to {\text{CH}}_{\text{3}}\text{COONa(aq)} \text{ + } {\text{H}}_{\text{2}}{\text{O(l) + CO}}_{\text{2}}\text{(g})$

Therefore, the spontaneous endothermic reaction is obtained as

${\text{NaHCO}}_{\text{3}}{\text{(s) + CH}}_{\text{3}}\text{COOH(aq)}\to {\text{CH}}_{\text{3}}{\text{COONa(aq) + H}}_{\text{2}}{\text{O(l) + CO}}_{\text{2}}\text{(g})$