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 for the given reaction, \(\mathrm{C}_{4} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g),\) is \(\Delta H = -158\ \mathrm{kJ/mol}\).

Step-by-step solution

Write the combustion reactions for the given compounds

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

Manipulate the combustion reactions to obtain the target reaction

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: $$ 2\mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \longrightarrow 2\mathrm{H}_{2}\mathrm{O}(l) \quad \Delta H_{3}^{'} = -572\ \mathrm{kJ/mol} $$

Add the manipulated reactions to get the target reaction and calculate \(\Delta H\)

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 \(\Delta H = -158\ \mathrm{kJ/mol}\).

Key concepts

Combustion Reactions

Combustion reactions are chemical reactions where a substance reacts with oxygen, releasing energy in the form of heat and light. This type of reaction is often associated with burning a substance, such as wood or gasoline. In chemistry, combustion reactions are particularly significant in understanding energy changes and chemical transformations. When compounds that contain carbon and hydrogen, known as hydrocarbons, undergo combustion, they typically produce carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)).

These reactions are exothermic, meaning they release energy. This is why combustion reactions are essential in engines and heating systems. For instance:

• The combustion of methane (\(\text{CH}_4\)) is: \[\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)\]
• The combustion of propane (\(\text{C}_3\text{H}_8\)) is: \[\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)\]

These reactions highlight the principle of conservation of mass where the number of each type of atom is the same before and after the reaction.

Hess's Law

Hess's Law is a fundamental principle in chemistry that states the total enthalpy change for a chemical reaction is the same, regardless of the number of stages or steps in which the reaction occurs. This means that if you can represent a chemical reaction as a series of steps, the sum of the enthalpy changes of these steps is equal to the enthalpy change of the reaction as a whole.

Hess's Law relies on the concept that enthalpy is a state function, which means its value depends only on the initial and final states of the system, not on the path taken to get from one to the other. This makes it extremely useful for calculating the enthalpy changes of reactions that are difficult to assess directly.

In practical terms:

• To find the enthalpy change (\(\Delta H\)) for a reaction, you can sum the enthalpy changes for individual steps that lead to the overall reaction.
• This is often shown using thermochemical equations.
• Hess's Law provides a way to construct difficult reactions from easier ones.

Using our example, we used Hess's Law to manipulate given combustion reactions to deduce the enthalpy change for a desired reaction.

Enthalpy of Combustion

The enthalpy of combustion is a specific type of enthalpy change (\(\Delta H\)) that occurs when one mole of a substance is completely burnt in oxygen. This value is usually expressed in kilojoules per mole (\(\text{kJ/mol}\)) and indicates how much energy is released during the combustion.

In practice, the enthalpy of combustion is important for understanding how energetic a substance is when used as a fuel. For example, the higher the enthalpy of combustion, the more energy is released when the fuel is burned, making it more efficient and useful.

It’s calculated from experiments or using tables of standard values, allowing chemists to predict the energy outputs of different reactions without performing arduous experimental measures each time. In our exercise, we are given the enthalpies of combustion for \(\text{C}_4\text{H}_4, \text{C}_4\text{H}_8\), and \(\text{H}_2\), which allowed us to use them to calculate the enthalpy change for a related reaction using Hess's Law.

Thermochemical Equations

Thermochemical equations are balanced chemical equations that include the enthalpy change, \((\Delta H)\), of the reaction, providing a clearer picture of the energy involved in chemical processes. These equations present both the substances involved and the energy changes, which are crucial for understanding how reactions transfer energy.

In these equations:

• The reactants and products are listed with their physical states.
• The enthalpy change (\(\Delta H\)) is included as part of the equation.
• An exothermic reaction will have a negative \(\Delta H\), indicating that energy is released.

Consider the combustion reaction of propane as an example: \[\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g); \Delta H = -2219 \text{ kJ/mol}\]
This thermochemical equation tells us not only what the reactants and products are, but also how much energy is released in the process. Such equations are vital for engineers and scientists when designing processes that involve chemical reactions, ensuring they can predict both the products and the energetic costs or outputs of these reactions.