Question

Consider the combustion of liquid methanol, CH₃OH(l):

CH₃OH(l) + 3/2 O₂(g) CO₂(g) + 2 H₂O(l) ∆H = –726.5 kJ

(a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is ∆H for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce H₂O(g) instead of H₂O(l), would you expect the magnitude of ∆H to increase, decrease, or stay the same? Explain.