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For the vaporization of ethanol (CH3CH2OH) at 25°C CH3CH2OH(l) → CH3CH2OH(g) ∆Hof(kJ mol-1) -277.63 -235.1 ∆Gof(kJ mol-1) -174.8 -168.6 a) What is the equilibrium constant at 25°C and what is the vapor pressure at 25°C? ∆Hof(kJ mol-1) : CH3CH2OH (l)= -277.63 CH3CH2OH (g)= -235.1 ∆Gof(kJ mol-1): CH3CH2OH (l)= -174.8 CH3CH2OH (g)= -168.6