wait…..the enthalpy of sabatier process is negative, while the reversed-reaction above is positive. so i think it is quite different.
if i increase the temperature, the equilibrium will shift to the right (product region). increasing pressure would not affect the equilibrium since the product...
it is known that the standard combustion enthalpy of methane is -882 kJ/mol
CH4 + 2 O2 --> CO2 + 2 H2O Hc = -882 kJ
based on hess's law, if a reaction is reversed, so does with the enthalpy
CO2 + 2 H2O --> CH4 + 2 O2 Hc = +882 kJ
which makes the reversed reaction become...