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 the reverse reaction

To find the enthalpy change for the reverse reaction, simply negate the enthalpy change for the forward reaction. If the forward reaction has an enthalpy change of ΔH = -726.5 kJ, the enthalpy change for the reverse reaction is: \[\Delta H_{reverse} = -(-726.5\, \text{kJ}) = 726.5\, \text{kJ}\]

02

(b) Balance the forward reaction and find new ΔH

To balance the forward reaction with whole-number coefficients, we need to ensure that there are equal numbers of atoms for each element on both sides of the equation. Currently, the equation is: \[\mathrm{CH}_{3}\mathrm{OH}(l) + \frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\] To avoid fractions, we can double the entire equation, resulting in the balanced equation: \[ 2\mathrm{CH}_{3}\mathrm{OH}(l) + 3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(l)\] Since the balanced equation is twice the original equation, the enthalpy change for this reaction is also twice as much. So the new enthalpy change is: \[\Delta H_{new} = 2(-726.5\, \text{kJ}) = -1453\, \text{kJ}\]

03

(c) Determine the thermodynamically favored reaction

A negative enthalpy change indicates that the reaction is exothermic, meaning it releases heat. Exothermic reactions are usually thermodynamically favored because they decrease the system's energy. In this case, the forward reaction has a negative enthalpy change, so it is more likely to be thermodynamically favored over the reverse reaction.

04

(d) Predict the effect of producing H2O(g) on ΔH

If the reaction were to produce H2O(g) instead of H2O(l), the enthalpy change would be expected to decrease (become less negative) because it takes energy to convert liquid water to water vapor. This energy is not accounted for in the original reaction enthalpy, so the enthalpy change for this new reaction would be: \[\Delta H_{new} \approx \Delta H_{forward} + \Delta H_{vaporization}\] where \(\Delta H_{forward}\) is the enthalpy change for the original forward reaction and \(\Delta H_{vaporization}\) is the energy required to vaporize water.

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