Calculate the standard free energy change for the below reaction at 25 degrees Celsius?

I'll assume these are all gases (in a coffee-cup calorimeter), although water liquid is possible in a bomb calorimeter. As a note, you should be especially careful that the values you look at are correct.

Two things that can give you trouble are:

So, you take the sum over the products, sum over the reactants, and subtract the sums, making sure you get the phases correct, and the coefficients match up.

Important standard state examples are:

(if you care, sulfur is actually orthorhombic in its standard state.) You should however definitely know that carbon graphite is the standard state of carbon, not diamond.)

Therefore, for

we just have:

To calculate the standard free energy change ($\Delta G^\circ$) for a reaction at 25 degrees Celsius, you can use the equation:

$\Delta G^\circ = \Delta H^\circ - T \cdot \Delta S^\circ$

Where:

• $\Delta H^\circ$ is the standard enthalpy change
• $\Delta S^\circ$ is the standard entropy change
• $T$ is the temperature in Kelvin

Without the reaction provided, it's not possible to calculate the standard free energy change. If you provide the reaction along with the standard enthalpy change ($\Delta H^\circ$) and the standard entropy change ($\Delta S^\circ$), I can assist you with the calculation.