Question

Calculate the value of $\Delta G$ for following reaction at 300 K.
$H_{2}O_{(s)}\longrightarrow H_{2}O_{(l)}$
$(\Delta H=7\text{ kJ}, \Delta S=24.8\text{ J K}^{-1})$

• A. $0.74\text{ kJ mol}^{-1}$
• B. $-0.82\text{ kJ mol}^{-1}$
• C. $0.21\text{ kJ mol}^{-1}$
• D. $-0.44\text{ kJ mol}^{-1}$

Hint:

Logic Tip: The process is ice melting to water. The normal melting point of ice is $0^\circ\text{C}$ (273 K). At any temperature above 273 K (like the 300 K given here), melting is spontaneous. A spontaneous process MUST have a negative $\Delta G$. This immediately rules out options A and C.

Answer 1

The Correct Option is D

Concept:
The change in Gibbs Free Energy ($\Delta G$) determines the spontaneity of a process at constant temperature and pressure. It is calculated using the Gibbs-Helmholtz equation: $$\Delta G = \Delta H - T\Delta S$$ where $\Delta H$ is the enthalpy change, $T$ is the absolute temperature in Kelvin, and $\Delta S$ is the entropy change. Care must be taken to ensure all energy units are consistent (either all Joules or all kilojoules).
Step 1: Identify and convert the given parameters into consistent units.
Enthalpy change, $\Delta H = 7\text{ kJ} = 7000\text{ J}$ Entropy change, $\Delta S = 24.8\text{ J K}^{-1}$ Temperature, $T = 300\text{ K}$ (Alternatively, convert $\Delta S$ to kilojoules: $\Delta S = 24.8 \times 10^{-3}\text{ kJ K}^{-1} = 0.0248\text{ kJ K}^{-1}$). Let's use kJ to match the options.
Step 2: Substitute values into the Gibbs-Helmholtz equation.
$$\Delta G = \Delta H - T\Delta S$$ $$\Delta G = 7\text{ kJ} - (300\text{ K} \times 0.0248\text{ kJ K}^{-1})$$
Step 3: Calculate the final $\Delta G$ value.
Calculate the $T\Delta S$ term first: $$T\Delta S = 300 \times 0.0248 = 3 \times 2.48 = 7.44\text{ kJ}$$ Now, subtract this from $\Delta H$: $$\Delta G = 7.00 - 7.44$$ $$\Delta G = -0.44\text{ kJ}$$ (The options present the units as $\text{kJ mol}^{-1}$, assuming the given values were per mole of water).