Consider the decomposition of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l),\) to gaseous acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g) :\) $$\mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \quad \Delta H=+630 \mathrm{kJ}$$ (a) What is the enthalpy change for the reverse reaction? (b) What is \(\Delta H\) for the formation of 1 mol of acetylene? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) were consumed instead of \(\mathrm{C}_{6} \mathrm{H}_{6}(l),\) 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

For the reverse reaction, we just need to reverse the sign of the enthalpy change. In this case, the reverse reaction would have an enthalpy change of \(-630~\text{kJ}\).

02

(b) Enthalpy change for the formation of 1 mol of acetylene

The given reaction shows the formation of 3 moles of acetylene and its enthalpy change. To find the enthalpy change for the formation of 1 mol of acetylene, we need to divide the given enthalpy change by 3: \[\Delta H_{\text{1 mol}} = \frac{630~\text{kJ}}{3} = 210~\text{kJ}\]

03

(c) Thermodynamically favored reaction

In general, exothermic reactions (with negative enthalpy changes) are more likely to be thermodynamically favored because they release energy into their surroundings. In this case, the forward reaction has a positive enthalpy change (+630 kJ), so it is an endothermic reaction, which means it absorbs energy from its surroundings. On the other hand, the reverse reaction, with an enthalpy change of -630 kJ, is exothermic. So, the reverse reaction is more likely to be thermodynamically favored.

04

(d) Effect of gaseous benzene on enthalpy change

The given enthalpy change (+630 kJ) is for the decomposition of liquid benzene. If gaseous benzene were consumed instead, the magnitude of the enthalpy change would likely decrease. The reasoning behind this is that converting a liquid to a gas typically involves breaking intermolecular forces, which requires energy input. Since liquid benzene becomes gaseous during the reaction, using gaseous benzene as the starting reactant would remove the need for the energy input required for vaporization. As a result, the overall enthalpy change for the reaction would likely decrease (but remain positive).

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