Question

Calculate the standard free energy change (∆G°) for the reaction: 2CH4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) at 298 K, given that the standard enthalpy change (∆H°) of the reaction is -890.4 kJ mol^-1 and the standard entropy change (∆S°) of the reaction is +292.2 J K^-1 mol^-1. Is the reaction spontaneous or non-spontaneous under standard conditions?

Answer 1

To calculate the standard free energy change  G  for the reaction, we can use the following equation:G = H - TSwhere H is the standard enthalpy change, T is the temperature in Kelvin, and S is the standard entropy change.Given values:H = -890.4 kJ mol^-1S = +292.2 J K^-1 mol^-1T = 298 KFirst, we need to convert the entropy change from J K^-1 mol^-1 to kJ K^-1 mol^-1:S = 292.2 J K^-1 mol^-1   1 kJ / 1000 J  = 0.2922 kJ K^-1 mol^-1Now, we can plug the values into the equation:G =  -890.4 kJ mol^-1  -  298 K  0.2922 kJ K^-1 mol^-1 G = -890.4 kJ mol^-1 - 87.0956 kJ mol^-1G = -977.4956 kJ mol^-1The standard free energy change for the reaction is approximately -977.5 kJ mol^-1.Since the G value is negative, the reaction is spontaneous under standard conditions.