Question

using the data from the previous steps, calculate the standard gibbs free energy reaction at 25 °c.
2al(s) + 3cl₂(g) → 2alcl₃(s)
δs°ᵣₓₙ = -507.3 j/k δh°ᵣₓₙ = -1408 kj
δg°ᵣₓₙ = ? kj
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for Gibbs Free Energy

The formula to calculate the standard Gibbs Free Energy change ($\Delta G^\circ_{\text{rxn}}$) 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, add 273.15:
$$T = 25 + 273.15 = 298.15\ \text{K}$$

Step3: Convert $\Delta S^\circ_{\text{rxn}}$ to kJ/K

Given $\Delta S^\circ_{\text{rxn}} = -507.3\ \text{J/K}$, convert to kJ/K (since $1\ \text{kJ} = 1000\ \text{J}$):
$$\Delta S^\circ_{\text{rxn}} = \frac{-507.3\ \text{J/K}}{1000} = -0.5073\ \text{kJ/K}$$

Step4: Substitute values into the formula

Substitute $\Delta H^\circ_{\text{rxn}} = -1408\ \text{kJ}$, $T = 298.15\ \text{K}$, and $\Delta S^\circ_{\text{rxn}} = -0.5073\ \text{kJ/K}$ into the formula:
$$\Delta G^\circ_{\text{rxn}} = -1408\ \text{kJ} - (298.15\ \text{K} \times -0.5073\ \text{kJ/K})$$

Step5: Calculate the product $T\Delta S^\circ_{\text{rxn}}$

First, compute $298.15 \times -0.5073$:
$$298.15 \times -0.5073 \approx -151.2$$
But since it’s $-T\Delta S^\circ_{\text{rxn}}$, the negative of that is:
$$- (298.15 \times -0.5073) \approx 151.2\ \text{kJ}$$

Step6: Add to $\Delta H^\circ_{\text{rxn}}$

Now, add this to $\Delta H^\circ_{\text{rxn}}$:
$$\Delta G^\circ_{\text{rxn}} = -1408\ \text{kJ} + 151.2\ \text{kJ} \approx -1256.8\ \text{kJ}$$

Answer:

-1257 (or -1256.8, depending on rounding; typically rounded to whole number)