X(g) + 3Y(g) ---- 2z(g) H = +ve. if the reaction above takes place at room temperature, the G will be

Analysis of the Question

This question tests your understanding of Gibbs Free Energy ($\Delta G$) and its relationship with Enthalpy ($\Delta H$), Entropy ($\Delta S$), and Temperature ($T$).

Concept Explanation

The relationship between these thermodynamic quantities is defined by the Gibbs-Helmholtz equation:

• $\Delta H$ (Enthalpy): The question states $\Delta H$ is positive ($+ve$), meaning the reaction is endothermic.
• $\Delta S$ (Entropy): We determine the sign of $\Delta S$ by looking at the change in the number of moles of gas. • Reactants: $1X + 3Y = 4$ moles of gas. • Products: $2Z = 2$ moles of gas. • Since the number of moles of gas decreases (from 4 to 2), the system becomes more ordered. Therefore, $\Delta S$ is negative ($-ve$).
• Temperature ($T$): The reaction occurs at room temperature.

Step-by-Step Solution

1. Substitute the signs into the Gibbs equation: $$\Delta G = (+ve) - T(-ve)$$
2. Simplifying the signs: $$\Delta G = (+ve) + (T \times +ve)$$
3. Since $T$ (absolute temperature in Kelvin) is always positive, the entire expression becomes: $$\Delta G = (+ve) + (+ve)$$
4. Adding two positive values always results in a positive value.
5. For a reaction to be spontaneous, $\Delta G$ must be negative. Since $\Delta G$ is positive, the reaction is non-spontaneous at room temperature.

Conclusion

Because the enthalpy change is positive (endothermic) and the entropy change is negative (decreasing disorder), the Gibbs free energy ($\Delta G$) will be positive at all temperatures, including room temperature.

Exam Tip

Remember the "Sign Rule" for spontaneity:

• If $\Delta H$ is negative and $\Delta S$ is positive, $\Delta G$ is always negative (Spontaneous).
• If $\Delta H$ is positive and $\Delta S$ is negative, $\Delta G$ is always positive (Non-spontaneous).
• If both are positive or both are negative, the spontaneity depends on the temperature!