Question

\( \Delta G \) for \( 3\text{X}_{(g)} + 2\text{Y}_{(g)} \rightarrow 3\text{Z}_{(g)} \) at 293 K is
( \( \Delta H^\circ = -13\,\text{kJ mol}^{-1} \), \( \Delta S^\circ = -45\,\text{J mol}^{-1}\text{K}^{-1} \) )

Hint:

The most common mistake in thermodynamics problems is adding Joules and kiloJoules directly. Always convert $\Delta S$ by dividing by 1000, or multiply $\Delta H$ by 1000 before proceeding with the subtraction.

Answer 1

Step 1: Write the Formula
\[ \Delta G = \Delta H - T\Delta S \]

Step 2: Convert Units (if required)
\[ \Delta H = -13 \text{ kJ/mol} = -13000 \text{ J/mol} \] \[ \Delta S = -45 \text{ J K}^{-1}\text{mol}^{-1}, \quad T = 293\,\text{K} \]

Step 3: Substitute Values
\[ \Delta G = -13000 - (293 \times -45) \]

Step 4: Calculate Entropy Term
\[ 293 \times (-45) = -13185 \text{ J/mol} \]

Step 5: Calculate \( \Delta G \)
\[ \Delta G = -13000 - (-13185) \] \[ \Delta G = -13000 + 13185 = 185 \text{ J/mol} \]

Step 6: Convert into kJ/mol
\[ \Delta G = 0.185 \text{ kJ/mol} \]

Step 7: Final Answer
\[ \boxed{\Delta G = +0.185 \text{ kJ/mol}} \]