Question

Trevor dissolves sodium hydroxide pellets in a beaker of water at room temperature, and notes that the beaker becomes warm. Which correctly designates the signs of ΔH, ΔS, and ΔG for this process?
A. ΔH > 0, ΔS > 0, and ΔG < 0
B. ΔH < 0, ΔS > 0, and ΔG < 0
C. ΔH > 0, ΔS > 0, and ΔG > 0
D. ΔH < 0, ΔS < 0, and ΔG > 0

Answer 1

Answer: B. ΔH < 0, ΔS > 0, and ΔG < 0

Explanation:

Exothermic reactions are those in which heat is released by the system and endothermic reactions are those in which heat is absorbed by the system.

$\Delta H$ for Exothermic reaction is negative and $\Delta H$ for Endothermic reaction is positive.

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

$\Delta S$ is positive when randomness increases and $\Delta S$ is negative when randomness decreases.

When NaOH is dissolved into water, the molecules change their state from solid to aqueous and thus randomness increases. The beaker becomes warm means the reaction is exothermic

$\Delta S$ is positive and $\Delta H$ is negative       Using Gibbs Helmholtz equation:

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

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

$\Delta G=-ve-ve$

Thus $\Delta G$ will be negative.

Thus $\Delta S$ is positive, $\Delta H$ is negative and $\Delta G$ will be negative.