Question

From the enthalpies of reaction $$ \begin{array}{rlr} \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g) & \Delta H=-537 \mathrm{~kJ} \\ \mathrm{C}(s)+2 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g) & \Delta H=-680 \mathrm{~kJ} \\ 2 \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g) & \Delta H=+52.3 \mathrm{~kJ} \end{array} $$ calculate \(\Delta H\) for the reaction of ethylene with \(\mathrm{F}_{2}\) : $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+6 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g) $$

Short answer

The enthalpy change for the reaction, \(\mathrm{C}_{2}\mathrm{H}_{4}(g) + 6 \mathrm{F}_{2}(g) \rightarrow 2 \mathrm{CF}_{4}(g) + 4 \mathrm{HF}(g)\), is \(\Delta H_4 = -2486.3 ~\mathrm{kJ}\).

Step-by-step solution

Write down the given reactions and the desired reaction.

The given reactions are: 1. \(\mathrm{H}_{2}(g) + \mathrm{F}_{2}(g) \rightarrow 2 \mathrm{HF}(g) \quad \Delta H_1 = -537 ~\mathrm{kJ}\) 2. \(\mathrm{C}(s) + 2 \mathrm{F}_{2}(g) \rightarrow \mathrm{CF}_{4}(g) \quad \Delta H_2 = -680 ~\mathrm{kJ}\) 3. \(2 \mathrm{C}(s) + 2 \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2}\mathrm{H}_{4}(g) \quad \Delta H_3 = +52.3 ~\mathrm{kJ}\) The desired reaction is: 4. \(\mathrm{C}_{2}\mathrm{H}_{4}(g) + 6 \mathrm{F}_{2}(g) \rightarrow 2 \mathrm{CF}_{4}(g) + 4 \mathrm{HF}(g) \quad \Delta H_4 = ?\)

Manipulate the given reactions to find a combination that matches the desired reaction.

Reverse reaction 3, which leads to: \( \mathrm{C}_{2}\mathrm{H}_{4}(g) \rightarrow 2 \mathrm{C}(s) + 2 \mathrm{H}_{2}(g) \quad -\Delta H_3= -52.3 ~\mathrm{kJ}\) Now multiply reaction 1 by 2, which leads to: \( 2 \mathrm{H}_{2}(g) + 2 \mathrm{F}_{2}(g) \rightarrow 4 \mathrm{HF}(g) \quad 2\Delta H_1 = -2 \times 537 ~\mathrm{kJ}= -1074 ~\mathrm{kJ}\) Lastly, multiply reaction 2 by 2, which leads to: \(2 \mathrm{C}(s) + 4 \mathrm{F}_{2}(g) \rightarrow 2 \mathrm{CF}_{4}(g) \quad 2\Delta H_2 = -2 \times 680 ~\mathrm{kJ} = -1360 ~\mathrm{kJ}\)

Add the manipulated reactions together and find the total enthalpy change.

Now we can add the manipulated reactions: \(\mathrm{C}_{2}\mathrm{H}_{4}(g) \rightarrow 2 \mathrm{C}(s) + 2 \mathrm{H}_{2}(g) \quad -52.3 ~\mathrm{kJ}\) \( 2 \mathrm{H}_{2}(g) + 2 \mathrm{F}_{2}(g) \rightarrow 4 \mathrm{HF}(g) \quad -1074 ~\mathrm{kJ}\) \(2 \mathrm{C}(s) + 4 \mathrm{F}_{2}(g) \rightarrow 2 \mathrm{CF}_{4}(g) \quad -1360 ~\mathrm{kJ}\) \(\mathrm{C}_{2}\mathrm{H}_{4}(g) + 6 \mathrm{F}_{2}(g) \rightarrow 2 \mathrm{CF}_{4}(g) + 4 \mathrm{HF}(g) \quad \Delta H_4\) The total enthalpy change for the desired reaction is: \(\Delta H_4 = -52.3 - 1074 - 1360 = -2486.3 ~\mathrm{kJ}\)

Write the final result.

The enthalpy change for the desired reaction, \(\mathrm{C}_{2}\mathrm{H}_{4}(g) + 6 \mathrm{F}_{2}(g) \rightarrow 2 \mathrm{CF}_{4}(g) + 4 \mathrm{HF}(g)\), is: \(\Delta H_4 = -2486.3 ~\mathrm{kJ}\)

Key concepts

Thermochemistry

Thermochemistry is the study of the energy and heat associated with chemical reactions and physical transformations. It is a branch of thermodynamics that focuses on the heat exchange accompanying the changes in state that matter undergoes, particularly during chemical reactions. At the heart of thermochemistry is the concept of enthalpy (symbolized as \( H \)), which is the total heat content of a system. Enthalpy changes, \( \Delta H \), indicate whether a reaction releases heat (exothermic, \( \Delta H < 0 \)) or absorbs heat (endothermic, \( \Delta H > 0 \)).

Understanding enthalpy changes is key to not just predicting whether a reaction will occur spontaneously but also to determining how much energy is involved. For students tackling homework problems involving thermochemistry, it's important to grasp that the enthalpy change of a reaction is equal to the energy transferred to or from the surroundings at constant pressure. This energy transfer can be measured or calculated, allowing us to quantify the energy changes within a chemical reaction.

Hess's Law

Hess's law is a principle that asserts the total enthalpy change during a chemical reaction is the same regardless of the number of steps the reaction is carried out in. This means that if you can't measure the enthalpy change directly, you can calculate it indirectly by using a series of reactions that, when added together, give you the overall reaction. Hess's law is grounded in the fact that enthalpy is a state function—its value is determined only by its current state, not the path the reaction takes to reach that state.

To apply Hess's law in practice, students may follow these steps: Write down the equations of the given reactions and manipulate them by reversing or scaling them so they add up to the desired equation. It's crucial to adjust the \( \Delta H \) values correspondingly. By algebraically summing the enthalpy changes of these manipulated reactions, students can calculate the enthalpy change for the desired reaction. In our exercise, we reversed and scaled reactions to deduce the enthalpy of ethylene reacting with fluorine, a clear application of Hess's law.

Chemical Reactions Enthalpy

When we talk about the enthalpy of chemical reactions, we are referring to the heat change that occurs during a reaction at constant pressure. This heat change, or reaction enthalpy, provides essential information about the energy efficiency and safety of chemical processes. For many chemical reactions, direct measurement of the enthalpy change isn't feasible. That's where calculation methods, like using Hess's law, come in handy.

Reaction enthalpies can be negative, indicating exothermic reactions that release heat into the surroundings, or positive for endothermic reactions that absorb heat. They allow us to predict not only the direction in which a reaction will proceed but also the extent to which it will occur. Having reliable values for reaction enthalpies contributes to our understanding of energy use, conservation, and sustainability within chemical engineering and environmental processes.