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 $$ \mathrm{C}_{4} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g) $$

We are given the enthalpies of combustion for C4H4, C4H8, and H2. Let's write their combustion reactions with the enthalpy values: $$ \mathrm{C}_{4} \mathrm{H}_{4}(g) + 4\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \quad \Delta H_{1} = -2341\ \mathrm{kJ/mol} $$ $$ \mathrm{C}_{4} \mathrm{H}_{8}(g) + 6\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(l) \quad \Delta H_{2} = -2755\ \mathrm{kJ/mol} $$ $$ \mathrm{H}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}\mathrm{O}(l) \quad \Delta H_{3} = -286\ \mathrm{kJ/mol} $$

Now, we need to manipulate these combustion reactions in such a way that when we add them up, we get our target reaction, which is: $$ \mathrm{C}_{4} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g) $$ First, we reverse the second reaction for C4H8: $$ 4\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g) + 6\mathrm{O}_{2}(g) \quad \Delta H_{2}^{'} = 2755\ \mathrm{kJ/mol} $$ Then, we multiply the third reaction (for H2) by 2: $$ 2\mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l) \quad \Delta H_{3}^{'} = -572\ \mathrm{kJ/mol} $$

Now we add the first reaction and the manipulated reactions for the second and third reactions. Notice that some terms will cancel out, and we'll be left with our target reaction: $$ \begin{aligned} \mathrm{C}_{4} \mathrm{H}_{4}(g) &+4 \mathrm{O}_{2}(g)+4 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \\ &\quad 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{C}_{4} \mathrm{H}_{8}(g)+6 \mathrm{O}_{2}(g) \end{aligned} $$ The target reaction is obtained after canceling out common terms on both sides: $$ \mathrm{C}_{4} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g) $$ Now we can find the enthalpy change \(\Delta H\) for the target reaction by adding the manipulated enthalpy values: $$ \Delta H = \Delta H_{1} + \Delta H_{2}^{'} + \Delta H_{3}^{'} = (-2341) + 2755 + (-572) = -158\ \mathrm{kJ/mol} $$ So, the enthalpy change for the given reaction is \(\Delta H = -158\ \mathrm{kJ/mol}\).