Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l) :\) $$\begin{aligned} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H &=-726.5 \mathrm{kJ} \end{aligned}$$ (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) 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 \(\mathrm{H}_{2} \mathrm{O}(g)\) instead of \(\mathrm{H}_{2} \mathrm{O}(l),\) would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

Step by step solution

01

Determining Enthalpy Change for Reverse Reaction

The reverse reaction of a chemical process is always opposite to the forward reaction. Therefore to find the enthalpy change for the reverse reaction, reverse the sign of the given enthalpy change. In this case, since the enthalpy change for the forward reaction is -726.5 kJ, the enthalpy change for the reverse reaction will be: ΔH_reverse = +726.5 kJ #b) Balance the reaction and find its enthalpy change#

02

Balancing the Equation

The given equation is already balanced with the lowest whole-number coefficients: CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l)

03

Finding Enthalpy Change

Since the equation is already balanced, the enthalpy change for the balanced forward reaction using whole-number coefficients is the same as what was given: ΔH = -726.5 kJ #c) Thermodynamically favored reaction#

04

Comparing Forward and Reverse Reactions

A forward reaction with a negative enthalpy change indicates that the reaction is exothermic and releases energy. In general, exothermic reactions are more thermodynamically favored in comparison to endothermic reactions (which have positive enthalpy changes). In this case, the forward reaction is exothermic with a negative value for ΔH, thus making it more likely to be thermodynamically favored over the reverse reaction. #d) Predicting change in enthalpy value if H2O was in its gaseous state#

05

Comparing Liquid and Gaseous H2O

If the reaction was written as: CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(g) We would need to figure out how the change in the phase of water will affect the enthalpy of the reaction. In general, when going from a liquid phase to a gaseous phase, the substance gains energy. As a result, the enthalpy change for the forward reaction should increase in magnitude (become less negative) to account for the additional energy needed to convert the water from the liquid to the gaseous phase. So, the magnitude of ΔH would be expected to increase.

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