Question

Answer the questions below by writing LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required) in the blanks. The reaction given below takes place in a cylinder that feels warm to the touch after the reaction is complete. \(\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \longrightarrow 2 \mathrm{AB}(s)\) (a) At all temperatures, \(\Delta S^{\circ}\) _____________ 0. (b) At all temperatures, \(\Delta H^{\circ}\) _____________ 0. (c) At all temperatures, \(\Delta G^{\circ}\) _____________ 0.

Short answer

Question: Based on the given reaction, determine the signs of ΔS°, ΔH°, and ΔG°. Answer: (a) ΔS° is LT (less than zero), (b) ΔH° is LT (less than zero), (c) ΔG° is LT (less than zero).

Step-by-step solution

(a) Determine ΔS°

To find the sign of the entropy change (ΔS°), we need to analyze the reactants and products in the reaction. In the given reaction: \(\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \longrightarrow 2 \mathrm{AB}(s)\). Notice that the reactants are both in the gaseous state, and the product is a solid. As a general rule, gases have higher entropy than solids, so the entropy decreases in this reaction. Therefore, ΔS° is less than zero (ΔS° < 0). The answer for (a) is LT.

(b) Determine ΔH°

The cylinder feels warm after the reaction, which indicates that it is an exothermic reaction. In an exothermic reaction, heat is released, meaning that ΔH° is negative (ΔH° < 0). The answer for (b) is LT.

(c) Determine ΔG°

To determine the sign of ΔG°, we can use the relationship between Gibbs free energy, enthalpy, and entropy. The equation is: ΔG° = ΔH° - TΔS°. Since both ΔH° and ΔS° are negative, and the temperature (T) is always positive, the value of TΔS° will be positive. Therefore, ΔG° = negative value - positive value, which means ΔG° is negative (ΔG° < 0) at all temperatures. The answer for (c) is LT.