Problem 65 - Chemistry: The Central Science in SI Units, Global Edition

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 for the target reaction, \(\mathrm{C}_{2}\mathrm{H}_{4}(g) + 6\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g) + 4 \mathrm{HF}(g)\), is \(\Delta H=-1806.3~\mathrm{kJ}\).

Step 1: Identify the main components of the reactions

Start by identifying the components present in each reaction and in the target reaction: - Reaction 1: Hydrogen gas, Fluorine gas, and Hydrogen Fluoride - Reaction 2: Carbon solid, Fluorine gas, and Carbon Tetrafluoride - Reaction 3: Carbon solid, Hydrogen gas, and Ethylene - Target reaction: Ethylene, Fluorine gas, Carbon Tetrafluoride, and Hydrogen Fluoride

Step 2: Determine a strategy to create the target reaction

Use the given reactions to obtain the desired reaction equation. Note that multiplying, dividing, or changing the direction of a reaction also alters the corresponding enthalpy change. We will perform the following modifications: 1. Reverse Reaction 3: We need ethylene as a reactant in the target reaction, so we will reverse the direction of reaction 3. 2. Multiply Reaction 1 by 2: In the target reaction, we require 4 moles of HF, so we need to multiply Reaction 1 by 2.

Step 3: Perform the modifications

Apply the modifications to the reactions and their corresponding enthalpies as follows: 1. Reverse Reaction 3: \(\mathrm{C}_{2}\mathrm{H}_{4}(g) \longrightarrow 2\mathrm{C}(s) + 2\mathrm{H}_{2}(g)\) - Change of direction, so \(\Delta H = -52.3~\mathrm{kJ}\) 2. Multiply Reaction 1 by 2: \(2[\mathrm{H}_{2}(g) + \mathrm{F}_{2}(g) \longrightarrow 2\mathrm{HF}(g)]\) - Multiply the equation by 2, so \(\Delta H = 2(-537) = -1074~\mathrm{kJ}\)

Step 4: Combine the modified reactions

Combine the modified reactions, including Reaction 2 (which remains unchanged), to form the target reaction: \(\mathrm{C}_{2}\mathrm{H}_{4}(g) \longrightarrow 2\mathrm{C}(s) + 2\mathrm{H}_{2}(g)\) with \(\Delta H = -52.3~\mathrm{kJ}\) \(2[\mathrm{H}_{2}(g) + \mathrm{F}_{2}(g) \longrightarrow 2\mathrm{HF}(g)]\) with \(\Delta H = -1074~\mathrm{kJ}\) \(\mathrm{C}(s) + 2\mathrm{F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g)\) with \(\Delta H = -680~\mathrm{kJ}\)

Step 5: Calculate the enthalpy change for the target reaction

When we add up the modified reactions, the left side of the equations will result in the left side of the target reaction and the right side of the equations will result in the right side of the target reaction. Add the corresponding enthalpies for each reaction to calculate the total enthalpy change for the target reaction: \(\Delta H_\text{target reaction}=-52.3~\mathrm{kJ}-1074~\mathrm{kJ}-680~\mathrm{kJ}=-1806.3~\mathrm{kJ}\) So, the enthalpy change for the target reaction, \(\mathrm{C}_{2}\mathrm{H}_{4}(g) + 6\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g) + 4 \mathrm{HF}(g)\), is \(\Delta H=-1806.3~\mathrm{kJ}\).

Key Concepts

Understanding Enthalpy Change

In thermochemistry, enthalpy change refers to the heat absorbed or released during a chemical reaction at constant pressure. It is represented by the symbol \( \Delta H \). The sign and magnitude of \( \Delta H \) tell us much about the reaction's thermal dynamics. - **Negative \( \Delta H \)** indicates a reaction that releases heat, known as an exothermic reaction.- **Positive \( \Delta H \)** indicates a reaction that absorbs heat, known as an endothermic reaction.Enthalpy change is a crucial concept because it helps us understand the energy flow within reactions. For example, in this problem, the calculated \( \Delta H \) for the target reaction is \(-1806.3 \text{ kJ} \), meaning the reaction releases a significant amount of heat, making it highly exothermic. Knowing how to manipulate and calculate enthalpy changes allows chemists to predict whether a reaction will be energetically favorable under set conditions.

Delving into Thermochemistry

Thermochemistry is the branch of chemistry that deals with the study of energy and heat involved in chemical and physical transformations. It provides insights into the energy changes that accompany chemical reactions. Key concepts include: - **Reaction Enthalpy**: Measures the total change in energy of a reaction. - **Hess's Law**: States that the total enthalpy change for a reaction is the same, regardless of the number of steps in which the reaction is carried out. Hess's Law is particularly useful in solving enthalpy problems like the one above, where multiple reactions are combined to get a target reaction. It allows us to manipulate given reactions (by reversing or scaling them) to create new reactions, providing a strategic way to calculate enthalpy changes when they cannot be measured directly.

Exploring Chemical Reactions

Chemical reactions involve rearrangements of atoms and molecules, leading to the transformation of reactants into products. These reactions can be described by chemical equations, which show the substances involved and their quantities. In the target reaction, ethylene \( \text{C}_2\text{H}_4 \) and fluorine gas \( \text{F}_2 \) react to form carbon tetrafluoride \( \text{CF}_4 \) and hydrogen fluoride \( \text{HF} \).When studying chemical reactions, consider:- **Reactants and Products**: Identify starting materials and the substances formed.- **Equation Manipulation**: Chemical equations can be reversed or adjusted by coefficients to achieve desired outcomes, impacting the enthalpy change.In the given example, understanding the interaction between ethylene and fluorine, and the relationship described in the reaction equations, is key to calculating the overall energy change. Through careful manipulation of individual reactions, we can predict products and energy changes effectively.