Question

using the data from the previous steps, calculate the standard gibbs free energy for the formation of rust at 25°c. 2fe(s) + 3h₂o(g) → fe₂o₃(s) + 3h₂(g) δs°ᵣₓₙ = -117 j/k δh°ᵣₓₙ = -140 kj δg°ᵣₓₙ = ? kj enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for Gibbs Free Energy

The formula relating standard Gibbs Free Energy ($\Delta G^\circ_{\text{rxn}}$), standard enthalpy change ($\Delta H^\circ_{\text{rxn}}$), standard entropy change ($\Delta S^\circ_{\text{rxn}}$), and temperature ($T$) is:
$$\Delta G^\circ_{\text{rxn}} = \Delta H^\circ_{\text{rxn}} - T\Delta S^\circ_{\text{rxn}}$$

Step2: Convert temperature to Kelvin

The temperature is $25^\circ\text{C}$. To convert to Kelvin:
$$T = 25 + 273.15 = 298.15\ \text{K}$$ (We can use $298\ \text{K}$ for simplicity.)

Step3: Convert entropy units to kJ/K

$\Delta S^\circ_{\text{rxn}} = -117\ \text{J/K}$. Convert to kJ/K:
$$\Delta S^\circ_{\text{rxn}} = -117\ \text{J/K} \times \frac{1\ \text{kJ}}{1000\ \text{J}} = -0.117\ \text{kJ/K}$$

Step4: Substitute values into the formula

$\Delta H^\circ_{\text{rxn}} = -140\ \text{kJ}$, $T = 298\ \text{K}$, $\Delta S^\circ_{\text{rxn}} = -0.117\ \text{kJ/K}$:
$$\Delta G^\circ_{\text{rxn}} = -140\ \text{kJ} - (298\ \text{K} \times -0.117\ \text{kJ/K})$$
First, calculate $298 \times -0.117$:
$$298 \times -0.117 = -34.866$$
Then, subtract this from $-140$:
$$\Delta G^\circ_{\text{rxn}} = -140 - (-34.866) = -140 + 34.866 = -105.134\ \text{kJ}$$ (Rounding to a reasonable precision, or using $298.15\ \text{K}$:
$$298.15 \times -0.117 = -34.88355$$
$$\Delta G^\circ_{\text{rxn}} = -140 - (298.15 \times -0.117) = -140 + 34.88355 \approx -105.116\ \text{kJ}$$)

Answer:

-105 (or more precise -105.1, depending on calculation; the key is the sign and magnitude. Using $T=298\ \text{K}$: $-140 - (298 \times -0.117) = -140 + 34.866 = -105.134 \approx -105$ or $-105.1$)