Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for \(\mathrm{C}_{4} \mathrm{H}_{4}(-2341 \mathrm{kJ} / \mathrm{mol}), \mathrm{C}_{4} \mathrm{H}_{8}\) \((-2755 \mathrm{kJ} / \mathrm{mol}),\) and \(\mathrm{H}_{2}(-286 \mathrm{kJ} / \mathrm{mol}),\) calculate \(\Delta H\) for the reaction $$\mathrm{C}_{4} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g)$$

Write the combustion reactions for \(\mathrm{C}_{4} \mathrm{H}_{4}\), \(\mathrm{H}_{2}\) and \(\mathrm{C}_{4} \mathrm{H}_{8}\) considering that the products are carbon dioxide (CO2) and water (H2O). 1. Combustion of \(\mathrm{C}_{4} \mathrm{H}_{4}\): $$\mathrm{C}_{4} \mathrm{H}_{4}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(g) \qquad \Delta H_{1} = -2341 \, \mathrm{kJ/mol}$$ 2. Combustion of \(\mathrm{H}_{2}\): $$\mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}\mathrm{O}(g) \qquad \Delta H_{2} = -286 \, \mathrm{kJ/mol}$$ 3. Combustion of \(\mathrm{C}_{4} \mathrm{H}_{8}\): $$\mathrm{C}_{4} \mathrm{H}_{8}(g) + 6 \mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(g) \qquad \Delta H_{3} = -2755 \, \mathrm{kJ/mol}$$

To calculate the change in enthalpy for the desired reaction \(\mathrm{C}_{4} \mathrm{H}_{4}(g) + 2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g)\), we need to rearrange the three given combustion reactions according to Hess's Law. 1. Reverse the first reaction, which will change the sign of \(\Delta H_{1}\): $$-[\mathrm{C}_{4} \mathrm{H}_{4}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(g)]$$ $$4\mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{4}(g) + 5 \mathrm{O}_{2}(g) \qquad \Delta H_{1}^{\prime} = 2341 \, \mathrm{kJ/mol}$$ 2. Multiply the second reaction by 2 to get 2 \(\mathrm{H}_{2}\) molecules: $$2[\mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}\mathrm{O}(g)]$$ $$2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(g) \qquad 2\Delta H_{2} = -572 \, \mathrm{kJ/mol}$$ 3. Reverse the third reaction, which will change the sign of \(\Delta H_{3}\): $$-[\mathrm{C}_{4} \mathrm{H}_{8}(g) + 6 \mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(g)]$$ $$4\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g) + 6 \mathrm{O}_{2}(g) \qquad \Delta H_{3}^{\prime} = 2755 \, \mathrm{kJ/mol}$$

Add the three modified reactions together and cancel out common terms on both sides: $$4\mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(g) + 2\mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) + 4\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{4}(g) + 5\mathrm{O}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(g) + \mathrm{C}_{4} \mathrm{H}_{8}(g) + 6\mathrm{O}_{2}(g)$$ Cancelling the terms: $$\mathrm{C}_{4} \mathrm{H}_{4}(g) + 2\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g)$$

Finally, sum the changes in enthalpy of the modified reactions to get the change in enthalpy for the desired reaction: $$\Delta H = \Delta H_{1}^{\prime} + 2\Delta H_{2} + \Delta H_{3}^{\prime} = 2341 - 572 + 2755 = \boxed{4524 \, \mathrm{kJ/mol}}$$