Questions: Consider the following reaction: I2(g) + Cl2(g) ⇌ 2 ICl(g) Kp=81.9 at 25°C. Calculate ΔGrxn for the reaction at 25°C under each of the following conditions. standard conditions Express your answer in kilojoules. ΔGrxn°= □ kJ

Consider the following reaction:
I2(g) + Cl2(g) ⇌ 2 ICl(g)
Kp=81.9 at 25°C.

Calculate ΔGrxn for the reaction at 25°C under each of the following conditions.
standard conditions
Express your answer in kilojoules.
ΔGrxn°=
□ kJ

Transcript text: Consider the following reaction: \[ \begin{array}{l} \mathrm{I}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ICl}(\mathrm{~g}) \\ K_{\mathrm{p}}=81.9 \text { at } 25^{\circ} \mathrm{C} . \end{array} \] Calculate $\Delta G_{r x n}$ for the reaction at $25^{\circ} \mathrm{C}$ under each of the following conditions. standard conditions Express your answer in kilojoules. \[ \Delta G_{\mathrm{rxn}}^{\circ}= \] $\square$ kJ

Solution

Solution Steps

Step 1: Understand the Relationship Between $ K_p $ and $\Delta G^\circ_{\text{rxn}}$

The relationship between the equilibrium constant $ K_p $ and the standard Gibbs free energy change $\Delta G^\circ_{\text{rxn}}$ is given by the equation:

\[ \Delta G^\circ_{\text{rxn}} = -RT \ln K_p \]

where:

$ R $ is the universal gas constant, $ R = 8.314 \, \text{J/mol} \cdot \text{K} $.
$ T $ is the temperature in Kelvin. For $ 25^\circ \text{C} $, $ T = 298.15 \, \text{K} $.
$ K_p $ is the equilibrium constant, given as 81.9.

Step 2: Convert Units and Calculate $\Delta G^\circ_{\text{rxn}}$

First, ensure all units are consistent. Since $ R $ is in J/mol·K, we need to convert the final answer to kJ/mol by dividing by 1000.

Substitute the values into the equation:

\[ \Delta G^\circ_{\text{rxn}} = - (8.314 \, \text{J/mol} \cdot \text{K}) \times (298.15 \, \text{K}) \times \ln(81.9) \]

Calculate $\ln(81.9)$:

\[ \ln(81.9) \approx 4.4053 \]

Now, calculate $\Delta G^\circ_{\text{rxn}}$:

\[ \Delta G^\circ_{\text{rxn}} = - (8.314) \times (298.15) \times 4.4053 \]

\[ \Delta G^\circ_{\text{rxn}} = -10920.5 \, \text{J/mol} \]

Convert to kJ/mol:

\[ \Delta G^\circ_{\text{rxn}} = -10.9205 \, \text{kJ/mol} \]