Problem 65 - Chemistry The Central Science

Question

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

Step-by-Step Solution

Answer

The enthalpy change, \(\Delta H\), for the reaction: $\mathrm{C}_{2} \mathrm{H}_{4}(g)+6 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g)$ is \(892.6\ \mathrm{kJ}\).

Step 1: Analyze the desired reaction

We need to manipulate the given reactions to create the desired reaction: $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+6 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g)$$

Step 2: Manipulate the given reactions

First, multiply the third reaction by 2: \[ 2(2 \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)): 4 \mathrm{C}(s)+4 \mathrm{H}_{2}(g) \longrightarrow 2\mathrm{C}_{2} \mathrm{H}_{4}(g) \quad \Delta H = 2(+52.3 \mathrm{kJ}) = 104.6 \mathrm{kJ} \] Now, multiply the first reaction by 4: \[ 4(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)): 4\mathrm{H}_{2}(g)+4\mathrm{F}_{2}(g) \longrightarrow 8 \mathrm{HF}(g) \quad \Delta H = 4(-537 \mathrm{kJ}) = -2148 \mathrm{kJ} \] Lastly, multiply the second reaction by 2: \[ 2(\mathrm{C}(s)+2 \mathrm{F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g)): 2\mathrm{C}(s)+4 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g) \quad \Delta H = 2(-680 \mathrm{kJ}) = -1360 \mathrm{kJ} \]

Step 3: Add and subtract the reactions

In order to get the desired reaction, add the modified second and third reactions and subtract the modified first reaction: $$ \begin{aligned} 4 \mathrm{C}(s)+4 \mathrm{H}_{2}(g) &\longrightarrow 2\mathrm{C}_{2} \mathrm{H}_{4}(g) &\quad \Delta H &= 104.6 \mathrm{kJ} \\ + 2\mathrm{C}(s)+4 \mathrm{F}_{2}(g) &\longrightarrow 2 \mathrm{CF}_{4}(g) &\quad \Delta H &= -1360 \mathrm{kJ} \\ - [4\mathrm{H}_{2}(g)+4\mathrm{F}_{2}(g) &\longrightarrow 8 \mathrm{HF}(g)] &\quad \Delta H &= 2148 \mathrm{kJ} \\ \hline \mathrm{C}_{2} \mathrm{H}_{4}(g)+6 \mathrm{F}_{2}(g) &\longrightarrow 2 \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g) &\quad \Delta H &= ? \end{aligned} $$

Step 4: Calculate the enthalpy change for the desired reaction

Add the enthalpy changes of the modified reactions to obtain the enthalpy change for the desired reaction: $$ \Delta H = 104.6\ \mathrm{kJ} - 1360\ \mathrm{kJ} + 2148\ \mathrm{kJ} = 892.6\ \mathrm{kJ} $$ Therefore, the enthalpy change of the reaction of ethylene with \(\mathrm{F}_{2}\) is: $$ \Delta H = 892.6\ \mathrm{kJ} $$

Key Concepts

Thermochemistry

Thermochemistry involves studying the energy and heat associated with chemical reactions and physical transformations. One key concept here is the system's enthalpy (\( H \)), which is a measurement of the total heat content. When a reaction occurs, the change in enthalpy, denoted as \( \Delta H \), can tell us if the process releases heat (exothermic, \( \Delta H < 0 \)) or absorbs heat (endothermic, \( \Delta H > 0 \)). For a given chemical equation, the enthalpy change corresponds to the difference between the enthalpy of the products and reactants, taking their stoichiometry into account.
Understanding the enthalpy change for a chemical reaction is crucial because it helps predict whether a reaction will occur spontaneously and what energy changes it will entail. For instance, the combustion of gasoline in an engine releases energy that propels a vehicle, while the endothermic nature of photosynthesis allows plants to absorb energy from sunlight.

Enthalpies of Reaction

Enthalpies of reaction refer specifically to the enthalpy change associated with a chemical reaction. This value can be measured directly using calorimetry or estimated using tabulated values known as standard enthalpies of formation. In the provided exercise, we are given the enthalpies of reaction for three different chemical processes. Each of these represents the amount of energy exchanged with the surroundings when the reaction proceeds as written, per mole of reaction.
For example, the reaction of hydrogen and fluorine to form hydrogen fluoride is highly exothermic, with \( \Delta H = -537 \text{kJ} \), indicating that a significant amount of energy is released as heat. This information is pivotal when calculating the enthalpy of a desired reaction that can be derived from these given reactions, like a puzzle that can be put together through Hess's Law.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves the use of balanced equations to determine the proportion of each substance involved. Stoichiometry is based on the conservation of mass and the principle that atoms are neither created nor destroyed in a chemical reaction. Therefore, we can manipulate stoichiometric coefficients to represent real-world quantities, like moles, and calculate corresponding enthalpies.
In the exercise, stoichiometry becomes key when adjusting the given reactions to align with the target reaction. The steps involve multiplying each reaction by coefficients to accurately represent the reaction of ethylene with fluorine. These coefficients ensure that all atoms accounted for in the reactants appear in the appropriate amounts among the products, allowing us to use Hess's Law to find the enthalpy change of the overall reaction.

Hess's Law

Hess's Law states that the total enthalpy change for a chemical reaction is the same, no matter how many steps or stages the reaction is carried out in. This principle allows us to calculate the enthalpy change for a complex reaction by breaking it down into a series of simpler reactions, whose enthalpy changes are known. These simpler reactions can be added or subtracted, as needed, to produce the overall reaction.
In the provided solution, Hess's Law was applied by manipulating and combining given reactions to obtain the target reaction. We are then able to sum the enthalpy changes of these manipulated reactions to determine the overall enthalpy change for the reaction of interest. It is crucial to ensure that all reactants and products cancel appropriately except for those appearing in the target equation. The accuracy of Hess's Law provides a powerful method to calculate enthalpy changes without the need for direct measurement.