The standard heats of combustion \(\left(\Delta H^{\circ}\right)\) per mole of 1,3-butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{g}) ;\) butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}) ;\) and \(\mathrm{H}_{2}(\mathrm{g})\) are \(-2540.2,-2877.6,\) and \(-285.8 \mathrm{kJ},\) respectively. Use these data to calculate the heat of hydrogenation of 1,3-butadiene to butane. $$\mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}) \quad \Delta H^{\circ}=?$$ [Hint: Write equations for the combustion reactions. In each combustion, the products are \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\left.\mathrm{H}_{2} \mathrm{O}(1) .\right]\)

Step by step solution

01

Write the combustion reactions

The combustion reactions of the reactants and product are as follows:\n\n1,3-Butadiene:\n\[C_{4}H_{6}(g) + 6O_{2}(g) \rightarrow 4CO_{2}(g) + 3H_{2}O(l)\] \[\Delta H_{1}^{\circ} = -2540.2 kJ\]\n\nHydrogen:\n\[H_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(l)\] \[\Delta H_{2}^{\circ} = -285.8 kJ\]\n\nButane:\n\[C_{4}H_{10}(g) + \frac{13}{2}O_{2}(g) \rightarrow 4CO_{2}(g) + 5H_{2}O(l)\] \[\Delta H_{3}^{\circ} = -2877.6 kJ\]

02

Apply Hess's Law

Hess's Law states that the total enthalpy change of a reaction is the same irrespective of the different reaction routes taken, provided the initial and final conditions are the same. In this problem, the hydrogenation reaction of 1,3-butadiene to butane will be calculated by rearranging the combustion reactions. The combustion reactions of 1,3-butadiene and hydrogen have to be summed to give the combustion reaction of butane. Therefore: \[\Delta H_{total}^{\circ} = \Delta H_{3}^{\circ} - \Delta H_{1}^{\circ} - 2\Delta H_{2}^{\circ}\]

03

Substitute the values

Substitute the given standard heats of combustion into the formula: \[\Delta H_{total}^{\circ} = -2877.6 kJ - (-2540.2 kJ) - 2(-285.8 kJ)\]

04

Calculate the heat of hydrogenation

Calculate the total enthalpy change to find the heat of hydrogenation: \[ \Delta H_{total}^{\circ} = -2877.6 kJ + 2540.2 kJ + 2 * 285.8 kJ = -119.6 kJ \]

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