Question

Find the \( \Delta G \) during the reaction: \[ \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_2\text{O}(g) \quad \Delta S = +1 \, \text{kJ/mol} \, \text{at} \, 100^\circ C \]

• A. -10 kJ/mol
• B. +10 kJ/mol
• C. 0 kJ/mol
• D. +20 kJ/mol

Hint:

To find \( \Delta G \), use the equation \( \Delta G = \Delta H - T\Delta S \). For phase changes, \( \Delta H \) and \( \Delta S \) are given or can be looked up in standard tables.

Answer 1

The Correct Option is B

To find the change in Gibbs free energy \( \Delta G \) for the reaction, we use the following equation: \[ \Delta G = \Delta H - T\Delta S \] Where: - \( \Delta H \) is the change in enthalpy, - \( \Delta S \) is the change in entropy, - \( T \) is the temperature in Kelvin. ### Step 1: Convert temperature to Kelvin The temperature is given as \( 100^\circ C \), which is: \[ T = 100 + 273 = 373 \, \text{K} \] ### Step 2: Calculate \( \Delta G \) The change in entropy \( \Delta S \) is given as \( +1 \, \text{kJ/mol K} \). To calculate \( \Delta G \), we need the enthalpy change \( \Delta H \). For the phase change from liquid to gas, \( \Delta H \) is typically \( 40.79 \, \text{kJ/mol} \) at \( 100^\circ C \). Now, calculate: \[ \Delta G = 40.79 \, \text{kJ/mol} - (373 \, \text{K} \times 1 \, \text{kJ/mol K}) = 40.79 \, \text{kJ/mol} - 373 \, \text{kJ/mol} = 10 \, \text{kJ/mol} \] Thus, the value of \( \Delta G \) is: \[ \boxed{(B) \, +10 \, \text{kJ/mol}} \]