Problem 104 - Chemistry: The Central Science

Question

5.104 We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) $$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$ \begin{array}{ll} \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-890.3 \mathrm{~kJ} \\ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) & \Delta H^{\circ}=-136.3 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-571.6 \mathrm{~kJ} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-3120.8 \mathrm{~kJ} \end{array} $$

Step-by-Step Solution

Answer

The standard enthalpy change for the conversion of methane to ethylene is \(\Delta H^{\circ} = -84 \,\text{kJ}\).

Step 1: Write out the target reaction

The target reaction is the conversion of methane to ethylene: \[ 2\,\text{CH}_{4}(g) \longrightarrow \text{C}_{2}\text{H}_{4}(g)+\text{H}_{2}(g) \]

Step 2: Rearrange and scale the given reactions

We have to rearrange and scale the given reactions in such a way that they result in the target reaction when added together. For the first reaction, we need \(\,\text{CH}_{4}(g)\) as a reactant, so keep this reaction as is, and multiply it by 2: \[ 2(\text{CH}_{4}(g) + 2\,\text{O}_{2}(g) \longrightarrow\text{CO}_{2}(g)+2\,\text{H}_{2}\text{O}(l)), \Delta H^{\circ}=2(-890.3\,\text{kJ}) \] For the second reaction, we need \(\,\text{C}_{2}\text{H}_{4}(g)\) as a product, so let's reverse the reaction and multiply by 1: \[ 1(\text{C}_{2}\text{H}_{6}(g) \longrightarrow\text{C}_{2}\text{H}_{4}(g)+\text{H}_{2}(g)), \Delta H^{\circ}=1(136.3\,\text{kJ}) \] For the third reaction, it can be completely eliminated because \(\text{H}_{2}\text{O}\) doesn't appear in the target reaction. For the fourth reaction, we need 4 \(\,\text{CO}_{2}(g)\), whereas it has only 1 \(\,\text{C}_{2}\text{H}_{4}(g)\) in it. We can reverse the reaction and divide by 2: \[ \frac{1}{2}(4\,\text{CO}_{2}(g)+6\,\text{H}_{2}\text{O}(l)\longrightarrow 2\,\text{C}_{2}\text{H}_{6}(g)+7\,\text{O}_{2}(g)), \Delta H^{\circ}=\frac{1}{2}(3120.8\,\text{kJ}) \] Now let's add these modified reactions.

Step 3: Add the modified reactions

Now, we'll add the modified reactions: \[ 2\,\text{CH}_{4}(g)+4\,\text{O}_{2}(g) \longrightarrow 4\,\text{CO}_{2}(g)+4\,\text{H}_{2}\text{O}(l) \\ \text{C}_{2}\text{H}_{6}(g) \longrightarrow\text{C}_{2}\text{H}_{4}(g)+\text{H}_{2}(g) \\ 4\,\text{CO}_{2}(g)+6\,\text{H}_{2}\text{O}(l) \longrightarrow 2\,\text{C}_{2}\text{H}_{6}(g)+7\,\text{O}_{2}(g) \] Adding these reactions gives us: \[ 2\,\text{CH}_{4}(g) + 4\,\text{O}_{2}(g) + \text{C}_{2}\text{H}_{6}(g) - 4\,\text{CO}_{2}(g) - 6\,\text{H}_{2}\text{O}(l) + 4\,\text{CO}_{2}(g) + 6\,\text{H}_{2}\text{O}(l) \longrightarrow \\ 4\,\text{CO}_{2}(g) + 4\,\text{H}_{2}\text{O}(l) + \text{C}_{2}\text{H}_{4}(g) + \text{H}_{2}(g) - 2\,\text{C}_{2}\text{H}_{6}(g) - 7\,\text{O}_{2}(g) \] Which simplifies to: \[ 2\,\text{CH}_{4}(g) \longrightarrow\text{C}_{2}\text{H}_{4}(g)+\text{H}_{2}(g) \]

Step 4: Calculate the standard enthalpy change of the target reaction

Now we sum the standard enthalpy changes of each modified reaction to calculate the standard enthalpy change for the target reaction: \(\Delta H^{\circ} = 2(-890.3\,\text{kJ}) + 1(136.3\,\text{kJ}) + \frac{1}{2}(3120.8\,\text{kJ})\) \(\Delta H^{\circ} = -1780.6\,\text{kJ} + 136.3\,\text{kJ} + 1560.4\,\text{kJ}\) \(\Delta H^{\circ} = -84\,\text{kJ}\) Therefore, the standard enthalpy change for the target reaction is \(\Delta H^{\circ} = -84 \,\text{kJ}\).

Key Concepts

enthalpy change

The enthalpy change, represented as \(\Delta H\), is a measure of the total energy change in a chemical reaction. Specifically, it represents the difference between the energy absorbed to break bonds in the reactants and the energy released when new bonds form in the products. An important detail is that enthalpy includes both the internal energy of the system and the energy required to make room for the system by displacing its surroundings.
When a reaction releases heat, it is termed exothermic, and the enthalpy change \(\Delta H\) is negative. Conversely, when a reaction absorbs heat, it is endothermic, and \(\Delta H\) is positive.
In the context of the exercise, Hess's Law is applied to compute the enthalpy change for a reaction that cannot be measured directly. Hess's Law asserts that the total enthalpy change is the same, regardless of whether the reaction occurs in one step or multiple steps. This property is what allows us to calculate the enthalpy change for complex reactions by using known data for simpler reactions.

thermochemical equations

Thermochemical equations are balanced chemical equations that also include the enthalpy change, \(\Delta H\), for the reaction. These equations provide specific information about the energy exchange associated with chemical reactions.
Unlike standard equations, a thermochemical equation gives insights into the heat absorbed or released during a reaction, ensuring a deeper understanding of the energy dynamics. For instance, when a reaction shows a negative \(\Delta H\), it means that the reaction releases energy to its surroundings, making it exothermic. Alternatively, a positive \(\Delta H\) indicates the reaction is endothermic, absorbing energy.
In the given exercise, thermochemical equations for several reactions were considered for rearrangement and manipulation in order to deduce the enthalpy change of a target reaction. Each step in manipulating these equations is underpinned by maintaining proper stoichiometry and energy balance.

chemical reactions

Chemical reactions are processes in which substances, known as reactants, are transformed into new substances, known as products. These transformations involve breaking and forming chemical bonds, which either require or release energy.
During a chemical reaction, the conservation of mass must occur; thus, the number of atoms of each element is consistent on both sides of the equation. Understanding chemical reactions requires recognizing the role of reactants, products, reaction conditions, and energy changes.
In the context of the exercise, understanding the detailed reaction between methane and other compounds to form ethylene is necessary. The objective is to effectively arrange data from known reactions into a sequence that precisely mimics the real-world process of the target reaction, as stipulated by Hess's Law.