For the reaction,
3C2H2(g) ===> C6H6 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 (in kJ) at 25°C.
What is the formula that relates stand. entropy, enthalpy and free energy changes?
Delta G= -631000J - (298.15)(-430J)
= -502795J
=-502.7955KJ
Did I do this right?
3C2H2 --> C6H6
The formula that relates standard entropy change (ΔS), standard enthalpy change (ΔH), and standard free energy change (ΔG) is:
ΔG = ΔH - TΔS
where:
ΔG is the standard free energy change
ΔH is the standard enthalpy change
ΔS is the standard entropy change
T is the temperature in Kelvin
To calculate the standard free energy change at 25°C, we need to convert the temperature from Celsius to Kelvin. The conversion formula is:
T (in Kelvin) = T (in Celsius) + 273.15
Using this formula, the temperature in Kelvin would be:
T = 25 + 273.15 = 298.15 K
Now we can substitute the values into the formula:
ΔG = ΔH - TΔS
ΔG = -631 kJ - (298.15 K)(-430 J/K)
Before we proceed with the calculation, we need to convert the units so that they are consistent. We can convert 430 J/K to kJ/K by dividing by 1000:
ΔG = -631 kJ - (298.15 K)(0.430 kJ/K)
Now we can perform the calculation:
ΔG = -631 kJ - 128.14 kJ
ΔG = -759.14 kJ
Therefore, the standard free energy change at 25°C for the given reaction is approximately -759.14 kJ.
That looks ok to me. You may want to check the number of significant figures.
delta G = delta H - TdeltaS .
Post your work if you need additional assistance.