The standard Gibbs free energy change ($\Delta G^\circ$) for a reaction can be calculated using the following formula:

$\Delta G^\circ = -RT \ln K$

where:

• $R$ is the gas constant, which is $8.314 \, \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$,
• $T$ is the temperature in Kelvin ($K$),
• $K$ is the equilibrium constant.

Step 1: Convert Temperature to Kelvin

Given $T = 840^\circ C$: $T = 840 + 273.15 = 1113.15 \, K$

Step 2: Substitute the Values

• $R = 8.314 \, \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$,
• $T = 1113.15 \, K$,
• $K = 3.3 \times 10^{-10}$.

Step 3: Calculate $\Delta G^\circ$

$\Delta G^\circ = - (8.314) (1113.15) \ln (3.3 \times 10^{-10})$

Now let's compute this numerically:

$\ln (3.3 \times 10^{-10}) \approx -21.83$

$\Delta G^\circ \approx - (8.314) (1113.15) (-21.83) \approx 202277.4 \, \text{J/mol} = 202.3 \, \text{kJ/mol}$

Final Answer:

$\Delta G^\circ \approx 202.3 \, \text{kJ/mol}$

Would you like further details or have any questions? Here are some related questions to explore:

1. How does a change in temperature affect the value of $\Delta G^\circ$?
2. What happens to $\Delta G^\circ$ if $K$ is increased?
3. What is the physical significance of a positive or negative $\Delta G^\circ$?
4. How does $\Delta G^\circ$ relate to the spontaneity of a reaction?
5. What other thermodynamic parameters can be used to describe chemical reactions?

Tip: Always ensure the temperature is in Kelvin when using thermodynamic equations like $\Delta G^\circ = -RT \ln K$.