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

(a) Enthalpy Change for Reverse Reaction

To find the enthalpy change for the reverse reaction, we simply change the sign of the given enthalpy change. The enthalpy change of the reverse reaction is the opposite of the forward reaction. In this case, the enthalpy change for the reverse reaction is: \(\Delta H_{\text{reverse}} = +726.5 \text{ kJ}\)

02

(b) Balancing the Reaction and Finding the Enthalpy Change

The given reaction equation is already balanced with whole-number coefficients: $$ \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ Since the reaction is balanced and the enthalpy change is already provided, the enthalpy change for the balanced reaction is the same: \(\Delta H = -726.5 \text{ kJ}\)

03

(c) Thermodynamically Favored Reaction

A reaction is thermodynamically favored if it proceeds with a decrease in enthalpy (negative \(\Delta H\)) since it releases energy to the surroundings. In this case, the forward reaction has a negative enthalpy change, which means it is thermodynamically favored over the reverse reaction.

04

(d) Effect of the Product Being Water Vapor Instead of Liquid Water

If the product were water vapor (\(\mathrm{H}_2\mathrm{O}(g)\)) instead of liquid water (\(\mathrm{H}_2\mathrm{O}(l)\)), we need to take into account the change in the phase of water. The enthalpy change for vaporization of water is positive since energy is required to convert liquid water to water vapor. Therefore, the overall enthalpy change of the reaction would be greater (less negative) than the given value. So, the magnitude of \(\Delta H\) would decrease (becoming less negative).

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