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) Calculate the enthalpy change for the reverse reaction

To find the enthalpy change for the reverse reaction, we simply change the sign of the given enthalpy change for the forward reaction. In this case, the enthalpy change for the reverse reaction is: \[-\Delta H = -(+630 \,\text{kJ}) = -630 \,\text{kJ}\]

02

(b) Calculate the enthalpy change for the formation of 1 mol of acetylene

We are given that the formation of 3 moles of acetylene has an enthalpy change of \(+630\) kJ. To find the enthalpy change for 1 mole of acetylene, we simply divide this value by 3: \[\Delta H_{1 \,\text{mol of } C_2H_2} = \frac{+630 \,\text{kJ}}{3} = +210 \,\text{kJ}\]

03

(c) Determine which reaction is thermodynamically favored

In general, a reaction with a negative enthalpy change (\(\Delta H<0\)) is exothermic and tends to be thermodynamically favored since it releases energy. A reaction with a positive enthalpy change (\(\Delta H>0\)) is endothermic and tends to be less thermodynamically favored since it absorbs energy. In this case, the forward reaction has a positive enthalpy change while the reverse reaction has a negative enthalpy change. Therefore, the reverse reaction is more likely to be thermodynamically favored.

04

(d) Expected change in enthalpy if gaseous benzene is consumed

Changing the phase of benzene from liquid to gas requires energy (known as the heat of vaporization). Therefore, if gaseous benzene were to be consumed instead of liquid benzene in the reaction, additional energy would be needed to turn liquid benzene into gaseous benzene before the reaction can occur. This would result in an increased magnitude of the enthalpy change (\(\Delta H\)) for the reaction.

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