for the reaction 2C2H2(g)=>C6H6(l) at 25 C, the standard enthalpy change is -631 kJ and the standard entropy change is -430 J/K. calculate the standard free energy change at 25 C. show all work

Help please!

You need to balance the equation which may explain why you can't get the answer you should.

3C2H2 --> C6H6
dG = dH-TdS
dG = -631,000 J - 298(-430) = ?

HFCSBZ

To calculate the standard free energy change (ΔG°) at 25°C, we can use the equation:

ΔG° = ΔH° - TΔS°

where:
- ΔH° is the standard enthalpy change
- T is the temperature in kelvin (25°C = 298 K)
- ΔS° is the standard entropy change

Given:
- ΔH° = -631 kJ (convert to J: -631 kJ = -631,000 J)
- ΔS° = -430 J/K
- T = 298 K

Let's plug in these values into the equation and calculate ΔG°:

ΔG° = (-631,000 J) - (298 K)(-430 J/K)
ΔG° = -631,000 J + 128,740 J
ΔG° = -502,260 J

Finally, we need to convert ΔG° from joules to kilojoules:
ΔG° = -502,260 J (convert to kJ: -502,260 J = -502.26 kJ)

Therefore, the standard free energy change at 25°C for the given reaction is -502.26 kJ.