Question

Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the following exothermic reaction. 2C2H6(g) 7O2(g)4CO2(g) 6H2O(g)

Answer 1

Answer: $\Delta H$ of the reaction will be negative, $\Delta S$ of the reaction will be positive and $\Delta G$ of the reaction will be negative.

Explanation:

Thermodynamic properties are enthalpy change $(\Delta H)$, entropy change $(\Delta S)$ and Gibbs free energy$(\Delta G)$

Exothermic reactions are defined as the reactions in which energy is released in the form of heat. The enthalpy change $(\Delta H)$ of the reaction comes out to be negative for this kind of reaction.

Entropy change is defined as the change in the measure of randomness in the reaction. It is represented as $(\Delta S)$. Randomness of gaseous particles is more than that of liquid which is further more than that of solids.

For the given exothermic reaction:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

As, number of gaseous particles on the product side is more than the number of gaseous particles on the reactant side. So, the entropy change is positive. Hence, $\Delta S$ is positive.

The above reaction is spontaneous. Thus, the Gibbs free energy will be negative.

For the given reaction:

• $\Delta H$ = -ve
• $\Delta S$ = +ve
• $\Delta G$ = -ve