Problem 65 - Chemistry The Central Science In Si Units

Question

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

Step-by-Step Solution

Answer

The enthalpy change \( \Delta H \) for the target reaction is \(-2486.3 \mathrm{~kJ}\).

Step 1: Write down the target reaction and equations

The final reaction we need the enthalpy for is \( \mathrm{C}_{2} \mathrm{H}_{4}(g)+6 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g) \). We have three given reactions: \( \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g) \), \( \mathrm{C}(s)+2 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g) \), and \( 2 \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g) \), with their respective enthalpies.

Step 2: Manipulate the given reactions

We need 4 HF molecules, so double the first reaction: \( 2\mathrm{H}_{2}(g)+2\mathrm{F}_{2}(g) \longrightarrow 4 \mathrm{HF}(g) \) with \( \Delta H = -1074 \mathrm{~kJ} \). The second reaction—\( \mathrm{C}(s)+2 \mathrm{F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g) \)—is used as given. Reverse the third reaction to form ethylene instead of decompose it: \( \mathrm{C}_{2}\mathrm{H}_{4}(g) \longrightarrow 2 \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \) with \( \Delta H = -52.3 \mathrm{~kJ} \).

Step 3: Combine reactions appropriately

Align the manipulated reactions to match all reactants and products of the target reaction: Add the reversed third reaction \( \mathrm{C}_{2} \mathrm{H}_{4}(g) \longrightarrow 2 \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \) and twice the second reaction \( 2 (\mathrm{C}(s)+2 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g)) \). Add these to the doubled first reaction to cancel out intermediate compounds, forming the target reaction.

Step 4: Calculate the total enthalpy change

Now add the enthalpies: \( -1074 \mathrm{~kJ} \) from the doubled first reaction, \( -680 \mathrm{~kJ} \times 2 = -1360 \mathrm{~kJ} \) from the doubled second reaction, and \( -52.3 \mathrm{~kJ} \) from the reversed third reaction. The total \( \Delta H \) for the target reaction is \( -1074 - 1360 - 52.3 = -2486.3 \mathrm{~kJ} \).

Key Concepts

Enthalpy of Reaction

The enthalpy of reaction, represented as \( \Delta H \), is a crucial concept in thermodynamics. It describes the heat change at constant pressure during a chemical reaction.
In essence, it allows us to understand whether a reaction releases heat, making it exothermic, or absorbs heat, making it endothermic. A negative \( \Delta H \) indicates an exothermic reaction, releasing energy into the surroundings, whereas a positive \( \Delta H \) suggests an endothermic reaction, where energy is absorbed from the surroundings.
In the context of our exercise, we're dealing with reactions involving hydrogen, fluorine, carbon, and ethylene. Each of these reactions comes with its own enthalpy value. To determine the enthalpy for a new reaction, namely that of ethylene reacting with fluorine, we use these given values as building blocks. By calculating the net \( \Delta H \), we can identify whether this new reaction is exothermic or endothermic.

Manipulation of Chemical Equations

Manipulating chemical equations is like rearranging puzzle pieces to form a picture. Here, the picture is your target reaction.
To apply Hess's Law effectively, the equations given must be altered to match the desired final reaction.
First, identify the molecules and their roles in each reaction. We needed four HF molecules in the final reaction, thus doubling the first reaction was necessary to achieve this quantity.

• We doubled the equation: \( 2 \mathrm{H}_{2}(g) + 2 \mathrm{F}_{2}(g) \rightarrow 4 \mathrm{HF}(g) \) with \( \Delta H = -1074 \; \text{kJ} \).

For the carbon tetrafluoride in the target reaction, we used the second reaction without modification.

• \( \mathrm{C}(s) + 2 \mathrm{F}_{2}(g) \rightarrow \mathrm{CF}_{4}(g) \) with \( \Delta H = -680 \; \text{kJ} \).

The last equation needed to be reversed to produce the reactant ethylene:

• \( \mathrm{C}_{2}\mathrm{H}_{4}(g) \rightarrow 2 \mathrm{C}(s) + 2 \mathrm{H}_{2}(g) \) with \( \Delta H = -52.3 \; \text{kJ} \).

Throughout this process, ensure that when modifying reactions (like reversing or multiplying), you make the necessary adjustments to their enthalpies.

Calculation of Enthalpy Change

Calculating the enthalpy change in a reaction involves summing the enthalpy changes from manipulated equations. Hess's Law states that the total enthalpy change is the sum of all changes, irrespective of the steps taken in the reaction pathway.
Let’s dive into the numbers:

• The enthalpy for the first equation, when doubled, is \(-1074 \; \text{kJ}\).
• Twice the second equation’s enthalpy is \( -680 \; \text{kJ} \times 2 = -1360 \; \text{kJ} \).
• Finally, reversing the third equation alters its enthalpy to \(-52.3 \; \text{kJ}\).

Adding these gives us \( -1074 - 1360 - 52.3 = -2486.3 \; \text{kJ} \), which is the total enthalpy change of the reaction between ethylene and fluorine.
This negates the need for direct measuring in a lab setting, illustrating how manipulating known reactions gets us to new thermodynamic insights. Through this careful calculation and manipulation, Hess’s Law empowers us to piece together comprehensive reactions from known data.