Question

For the reaction:

$3Fe_{(s)} + 2O_2{(g)} \rightarrow Fe_3O_4{(s)}$

$\Delta H = -1650\,\text{kJ mol}^{-1}$, $\Delta S = -600\,\text{J K}^{-1} \text{mol}^{-1}$ at $300\,\text{K}$. What is the value of free energy change for the reaction at $300\,\text{K}$?

• A. $-1470$ J mol^-1
• B. $-1830$ J mol^-1
• C. $-147.02$ kJ mol^-1
• D. $-1830$ kJ mol^-1
• E. $-1470$ kJ mol^-1

Hint:

The Gibbs free energy change determines spontaneity: $\Delta G<0$ means a spontaneous reaction.

Answer 1

Answer: (E)

Step 1: The Gibbs free energy change is calculated using: \[ \Delta G = \Delta H - T \Delta S \] Step 2: Given: \[ \Delta H = -1650 { kJ mol}^{-1}, \quad \Delta S = -600 { J K}^{-1} {mol}^{-1} = -0.6 { kJ K}^{-1} {mol}^{-1}, \quad T = 300 { K} \] Step 3: Compute $\Delta G$: \[ \Delta G = -1650 - (300 \times -0.6) \] \[ \Delta G = -1650 + 180 \] \[ \Delta G = -1470 { kJ mol}^{-1} \] Step 4: Therefore, the correct answer is (E). \bigskip