Question

In the following reaction, the values of \( \Delta H \) and \( \Delta S \) at temperature 25 °C are -13.7 kcal/mole and -16.0 cal/(K·mole), respectively.
The value of \( \Delta G \) (in kcal/mole) of the reaction, rounded off to TWO decimal places, is ...............

Hint:

To calculate \( \Delta G \), ensure that all units are consistent, typically using kcal and K for thermodynamic calculations.

Answer 1

Correct Answer: -8.9 - -8.95

Step 1: Use the Gibbs free energy equation. 
The change in Gibbs free energy \( \Delta G \) is given by the equation: \[ \Delta G = \Delta H - T \Delta S, \] where \( \Delta H \) is the enthalpy change, \( \Delta S \) is the entropy change, and \( T \) is the temperature in Kelvin.

Step 2: Converting the values to consistent units. 
- \( \Delta H = -13.7 \, \text{kcal/mole} \) 

- \( \Delta S = -16.0 \ \text{cal}\,\mathrm{K^{-1}\,mol^{-1}} = -0.016 \ \text{kcal}\,\mathrm{K^{-1}\,mol^{-1}} \) 

- \( T = 25^\circ \text{C} = 298 \, \text{K} \)

Step 3: Substituting into the equation. 
\[ \Delta G = -13.7 - (298)(-0.016), \] \[ \Delta G = -13.7 + 4.768 = -8.932 \, \text{kcal/mole}. \]

Step 4: Conclusion. 
The value of \( \Delta G \) is \( \boxed{-8.93} \, \text{kcal/mole} \).