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Chapter 5: Problem 65

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 \(\mathrm{F}_{2}\) : $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+6 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g) $$

Short Answer

Expert verified
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 by step solution

01

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
02

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.
03

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 \(\Delta H = 2(-537) = -1074~\mathrm{kJ}\)
04

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 \(\Delta H = -680~\mathrm{kJ}\)
05

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 \(\Delta H=-1806.3~\mathrm{kJ}\).

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Most popular questions from this chapter

The standard enthalpies of formation of gaseous propyne $\left(\mathrm{C}_{3} \mathrm{H}_{4}\right),\( propylene \)\left(\mathrm{C}_{3} \mathrm{H}_{6}\right),\( and propane \)\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\( are \)+185.4,+20.4,\( and \)-103.8 \mathrm{~kJ} / \mathrm{mol}$, respectively. (a) Calculate the heat evolved per mole on combustion of each substance to yield \(\mathrm{CO}_{2}(g)\) and $\mathrm{H}_{2} \mathrm{O}(g) .\( (b) Calculate the heat evolved on combustion of \)1 \mathrm{~kg}\( of each substance. \)(\mathbf{c})$ Which is the most efficient fuel in terms of heat evolved per unit mass?

(a) What is the electrostatic potential energy (in joules) between an electron and a proton that are separated by \(230 \mathrm{pm}\) ? (b) What is the change in potential energy if the distance separating the electron and proton is increased to \(1.0 \mathrm{nm}\) ? (c) Does the potential energy of the two particles increase or decrease when the distance is increased to \(1.0 \mathrm{nm}\) ?

(a) Write an equation that expresses the first law of thermodynamics in terms of heat and work. (b) Under what conditions will the quantities \(q\) and \(w\) be negative numbers?

Suppose an Olympic diver who weighs \(52.0 \mathrm{~kg}\) executes a straight dive from a 10 -m platform. At the apex of the dive, the diver is $10.8 \mathrm{~m}$ above the surface of the water. (a) What is the potential energy of the diver at the apex of the dive, relative to the surface of the water? (b) Assuming that all the potential energy of the diver is converted into kinetic energy at the surface of the water, at what speed, in $\mathrm{m} / \mathrm{s}$, will the diver enter the water? (c) Does the diver do work on entering the water? Explain.

Butane \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) boils at $-0.5^{\circ} \mathrm{C} ;\( at this temperature it has a density of \)0.60 \mathrm{~g} / \mathrm{cm}^{3}\(. The enthalpy of formation of \)\mathrm{C}_{4} \mathrm{H}_{10}(g)\( is \)-124.7 \mathrm{~kJ} / \mathrm{mol},$ and the enthalpy of vaporiza- tion of \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) is $22.44 \mathrm{~kJ} / \mathrm{mol} .\( Calculate the enthalpy change when \)1 \mathrm{~L}$ of liquid \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) is burned in air to give \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) How does this compare with \(\Delta H\) for the complete combustion of \(1 \mathrm{~L}\) of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l) ?\) For $\mathrm{CH}_{3} \mathrm{OH}(l),\( the density at \)25^{\circ} \mathrm{C}\( is \)0.792 \mathrm{~g} / \mathrm{cm}^{3},\( and \)\Delta H_{f}^{\circ}=-239 \mathrm{~kJ} / \mathrm{mol}$.
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