Question

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

Short answer

The enthalpy change (\(\Delta H\)) for the reaction: $$\mathrm{C}_{4}\mathrm{H}_{4}(g) + 2\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4}\mathrm{H}_{8}(g)$$ is \(-842 \mathrm{k}\mathrm{J}/\mathrm{mol}\).

Step-by-step solution

Write balance the combustion reactions

The balanced combustion reactions for the given compounds are: $$\mathrm{C}_{4}\mathrm{H}_{4}(g) + 5\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \qquad \Delta H_1 = -2341 \mathrm{k}\mathrm{J}/\mathrm{mol}$$ $$\mathrm{C}_{4}\mathrm{H}_{8}(g) + 6\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(l) \qquad \Delta H_2 = -2755 \mathrm{k}\mathrm{J}/\mathrm{mol}$$ $$\mathrm{H}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}\mathrm{O}(l) \qquad \Delta H_3 = -286 \mathrm{k}\mathrm{J}/\mathrm{mol}$$

Manipulate the combustion reactions to match the desired reaction

The goal is to obtain the desired reaction: $$\mathrm{C}_{4}\mathrm{H}_{4}(g) + 2\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4}\mathrm{H}_{8}(g)$$ To do this, we need to manipulate the combustion reactions and then add or subtract them to get the desired reaction. First, reverse the first combustion reaction and add it to the second reaction: $$\mathrm{C}_{4}\mathrm{H}_{4}(g) + 4\mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 5\mathrm{O}_{2}(g) \qquad -\Delta H_1 = 2341 \mathrm{k}\mathrm{J}/\mathrm{mol}$$ $$+$$ $$\mathrm{C}_{4}\mathrm{H}_{8}(g) + 6\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2}\mathrm{O}(l) \qquad \Delta H_2 = -2755 \mathrm{k}\mathrm{J}/\mathrm{mol}$$ $$\Rightarrow$$ $$\mathrm{C}_{4}\mathrm{H}_{4}(g) + 6\mathrm{O}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{C}_{4}\mathrm{H}_{8}(g) + \mathrm{O}_{2}(g)$$ We are still missing \(2\mathrm{H}_{2}(g)\) on the reactant side. So, we will subtract two times the third combustion reaction (multiplied by 2): $$-\underbrace{2}_{\times 2}\left[\mathrm{H}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}\mathrm{O}(l)\right] \qquad -2\Delta H_3 = 572 \mathrm{k}\mathrm{J}/\mathrm{mol}$$ Finally, we have: $$\mathrm{C}_{4}\mathrm{H}_{4}(g) + 6\mathrm{O}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{C}_{4}\mathrm{H}_{8}(g) + \mathrm{O}_{2}(g)$$ $$-$$ $$2\mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l)$$ $$\Rightarrow$$ $$\mathrm{C}_{4}\mathrm{H}_{4}(g) + 2\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4}\mathrm{H}_{8}(g)$$

Add the enthalpies

Now we can add up the enthalpies from each manipulation to find the \(\Delta H\) for the desired reaction: $$\Delta H = -\Delta H_1 + \Delta H_2 - 2\Delta H_3$$ $$\Rightarrow \Delta H = 2341 - 2755 + 572 = -842 \mathrm{k}\mathrm{J}/\mathrm{mol}$$ So, the enthalpy change (\(\Delta H\)) for the reaction: $$\mathrm{C}_{4}\mathrm{H}_{4}(g) + 2\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4}\mathrm{H}_{8}(g)$$ is \(\boldsymbol{-842 \mathrm{k}\mathrm{J}/\mathrm{mol}}\).

Key concepts

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance combines with oxygen and releases heat. In the context of hydrocarbons like butene (\( \text{C}_4\text{H}_8 \text{g} \) and butadiene (\( \text{C}_4\text{H}_4 \text{g} \) ), the reaction typically produces carbon dioxide (\( \text{CO}_2 \) ) and water (\( \text{H}_2\text{O} \) ) as the primary products.

The combustion of hydrocarbons is a crucial process both in nature and industry. It's essential to understand because it not only powers vehicles and generates electricity but also is foundational in chemical thermodynamics. To fully grasp these reactions, it's important to visualize the compounds exchanging atoms with oxygen, resulting in an exothermic reaction that releases energy.

Enthalpy of Combustion

The enthalpy of combustion (\( \text{ΔH}_{\text{combustion}} \) ) is a term referring to the amount of heat released when one mole of a substance burns completely in oxygen under standard conditions. It's a specific case of enthalpy change reflecting how much energy is given off during a combustion reaction.

By using the known enthalpy of combustion for the substances involved, students can calculate the enthalpy change for related reactions. It's critical to remember that combustion is always exothermic, meaning that the sign of the enthalpy change is negative, indicating energy is being released to the surroundings.

Chemical Thermodynamics

Chemical thermodynamics deals with the study of energy changes accompanying chemical reactions and the direction in which these reactions can naturally proceed. The two primary concepts here are enthalpy (\( \text{H} \) ) which is the heat content and entropy (\( \text{S} \) ) which is a measure of disorder or randomness.

In the context of chemical reactions, the first and second laws of thermodynamics guide us to understand that energy cannot be created or destroyed, only converted, and that the universe tends toward increased entropy respectively. These principles help predict whether a reaction will occur spontaneously and what energy changes to expect, which is essential when calculating the enthalpy change of reactions, including combustions.

Balanced Chemical Equations

A balanced chemical equation reflects the law of conservation of mass by showing the same number of each type of atom on both the reactant and product sides. Maintaining this balance can become more complex when reactions involve multiple steps or require intermediate calculations, as demonstrated in our example.

One must be meticulous when manipulating equations to reflect the actual chemistry occurring, which is why reversing, adding, and subtracting combustion reactions is necessary to derive a new reaction's enthalpy change. It emphasizes the need for correct stoichiometry to accurately calculate energy changes such as the enthalpy change of a reaction.