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We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: $$2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)$$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$\begin{array}{ll}{\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-890.3 \mathrm{kJ}} \\ {\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)} & {\Delta H^{\circ}=-136.3 \mathrm{kJ}} \\ {2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-571.6 \mathrm{kJ}} \\ {2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-3120.8 \mathrm{kJ}}\end{array}$$

Short Answer

Expert verified
The enthalpy change for the reaction \(2 CH_4(g) \longrightarrow C_2H_4(g) + H_2(g)\) is \(\Delta H^\circ = -3479.3 kJ\).

Step by step solution

01

Analyze the given reactions

We have the following reactions and their enthalpy changes: 1. \(CH_4(g) + 2 O_2(g) \longrightarrow CO_2(g) + 2 H_2O(l)\), \(\Delta H_1^\circ = -890.3 kJ\) 2. \(C_2H_4(g) + H_2(g) \longrightarrow C_2H_6(g)\), \(\Delta H_2^\circ = -136.3 kJ\) 3. \(2 H_2(g) + O_2(g) \longrightarrow 2 H_2O(l)\), \(\Delta H_3^\circ = -571.6 kJ\) 4. \(2 C_2H_6(g) + 7 O_2(g) \longrightarrow 4 CO_2(g) + 6 H_2O(l)\), \(\Delta H_4^\circ = -3120.8 kJ\)
02

Manipulate reactions to form the target reaction

We want to form the reaction: \[2 CH_4(g) \longrightarrow C_2H_4(g) + H_2(g) \] Manipulate the given reactions as follows: - Multiply reaction 1 by 2: \[2 CH_4(g) + 4 O_2(g) \longrightarrow 2 CO_2(g) + 4 H_2O(l), \Delta H_{1'}^\circ = -1780.6 kJ\] - Reverse reaction 2: \[C_2H_6(g) \longrightarrow C_2H_4(g) + H_2(g), \Delta H_{2'}^\circ = 136.3 kJ\] - Multiply reaction 3 by 1/2 and reverse it: \[H_2O(l) \longrightarrow H_2(g) + 1/2 O_2(g), \Delta H_{3'}^\circ = 285.8 kJ\] - Use reaction 4 as it is: \[2 C_2H_6(g) + 7 O_2(g) \longrightarrow 4 CO_2(g) + 6 H_2O(l), \Delta H_4^\circ = -3120.8 kJ\]
03

Add the manipulated reactions to obtain the target reaction

Now, add the manipulated reactions together: 1. \(2 CH_4(g) + 4 O_2(g) \longrightarrow 2 CO_2(g) + 4 H_2O(l)\) 2. \(C_2H_6(g) \longrightarrow C_2H_4(g) + H_2(g) \) 3. \(H_2O(l) \longrightarrow H_2(g) + 1/2 O_2(g) \) 4. \(2 C_2H_6(g) + 7 O_2(g) \longrightarrow 4 CO_2(g) + 6 H_2O(l)\) Sum: \(2 CH_4(g) + 4 CO_2(g) + 4 H_2O(l) \longrightarrow C_2H_4(g) + H_2(g) + 2 CO_2(g) + 6 H_2O(l) + 4 CO_2(g) \) Simplify: \(2 CH_4(g) \longrightarrow C_2H_4(g) + H_2(g)\)
04

Calculate the enthalpy change for the target reaction

Now, sum the \(\Delta H^\circ\) values of the manipulated reactions: \[\Delta H = \Delta H_{1'}^\circ + \Delta H_{2'}^\circ + \Delta H_{3'}^\circ + \Delta H_{4}^\circ\] \[\Delta H = (-1780.6 kJ) + (136.3 kJ) + (285.8 kJ) + (-3120.8 kJ)\] \[\Delta H = -3479.3 kJ\] So, the enthalpy change for the reaction \(2 CH_4(g) \longrightarrow C_2H_4(g) + H_2(g)\) is \(\Delta H^\circ = -3479.3 kJ\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
Enthalpy change is an essential concept in chemistry that helps measure the heat absorbed or released during a chemical reaction. It is symbolized as \(\Delta H\). When a reaction occurs at constant pressure, the change in enthalpy represents the heat exchange with the surroundings.Several factors can influence the enthalpy change of a reaction:
  • The nature of reactants and products
  • The conditions under which the reaction occurs, such as pressure and temperature
  • The states of the substances involved (solid, liquid, gas)
The enthalpy change of a reaction can be either positive or negative:
  • A positive \(\Delta H\) indicates an endothermic reaction where heat is absorbed.
  • A negative \(\Delta H\) indicates an exothermic reaction where heat is released.
Understanding the enthalpy change helps in predicting whether a reaction requires energy input or releases energy, which is crucial for various industrial and laboratory processes.
Reaction Manipulation
Reaction manipulation involves adjusting chemical equations to help calculate unknown enthalpy changes using known data. According to Hess's Law, the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction pathway.Here are some common manipulation strategies:
  • Reversing reactions: Changing the direction of a reaction will change the sign of \(\Delta H\). For example, if a reaction is exothermic with \(\Delta H = -100 \, \text{kJ}\), the reverse reaction will be endothermic with \(\Delta H = +100 \, \text{kJ}\).
  • Multiplying reactions: If a reaction is multiplied by a coefficient, \(\Delta H\) is also multiplied by the same factor.
  • Adding reactions: Reactions can be added together, and their enthalpy changes are summed to find the overall \(\Delta H\).
By strategically manipulating known reactions, you can derive the enthalpy change of a target reaction that might be otherwise difficult to measure directly.
Thermochemical Equations
Thermochemical equations are balanced chemical equations that include the enthalpy change. They are a powerful way to represent the energy changes associated with chemical reactions, providing insight into whether energy is absorbed or released.Key features of thermochemical equations:
  • The balanced chemical equation shows the stoichiometry of the reactants and products.
  • The enthalpy change \(\Delta H\) is included, usually in kilojoules (\(\text{kJ}\)), and gives a thermodynamic perspective of the reaction.
  • States of matter (solid, liquid, gas) for each component are often specified, which is important as enthalpy changes can depend on these states.
When writing or interpreting thermochemical equations, it is crucial to remember:
  • Ensure the equation is balanced for both mass and charge.
  • The specified \(\Delta H\) is accurate under the given conditions.
  • Consider how changes in conditions (such as temperature or pressure) might affect the \(\Delta H\).
Thermochemical equations are a foundational tool in thermodynamics, helping chemists and engineers predict reaction behaviors and design energy-efficient processes.

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Most popular questions from this chapter

Without doing any calculations, predict the sign of \(\Delta H\) for each of the following reactions: $$\begin{array}{l}{\text { (a) } 2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)} \\ {\text { (b) } 2 \mathrm{F}(g) \longrightarrow \mathrm{F}_{2}(g)} \\ {\text { (c) } \mathrm{Mg}^{2+}(g)+2 \mathrm{Cl}^{-}(g) \longrightarrow \mathrm{MgCl}_{2}(s)} \\ {\text { (d) } \mathrm{HBr}(g) \longrightarrow \mathrm{H}(g)+\mathrm{Br}(g)}\end{array}$$

Ozone, \(\mathrm{O}_{3}(g),\) is a form of elemental oxygen that plays an important role in the absorption of ultraviolet radiation in the stratosphere. It decomposes to \(\mathrm{O}_{2}(g)\) at room temperature and pressure according to the following reaction: $$2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) \quad \Delta H=-284.6 \mathrm{kJ}$$ (a) What is the enthalpy change for this reaction per mole of \(\mathrm{O}_{3}(g) ?\) (b) Which has the higher enthalpy under these conditions, 2 \(\mathrm{O}_{3}(g)\) or 3 \(\mathrm{O}_{2}(g) ?\)

Consider two solutions, the first being 50.0 \(\mathrm{mL}\) of 1.00 \(\mathrm{MCuSO}_{4}\) and the second 50.0 \(\mathrm{mL}\) of 2.00 \(\mathrm{M} \mathrm{KOH}\) . When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 to \(27.7^{\circ} \mathrm{C}\) (a) Before mixing, how many grams of Cu are present in the solution of \(\mathrm{CuSO}_{4}\) ? (b) Predict the identity of the precipitate in the reaction. (c) Write complete and net ionic equations for the reaction that occurs when the two solutions are mixed. (d) From the calorimetric data, calculate \(\Delta H\) for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 \(\mathrm{mL}\) , and that the specific heat and density of the solution after mixing are the same as those of pure water.

Assume that the following reaction occurs at constant pressure: $$2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s)$$ (a) If you are given \(\Delta H\) for the reaction, what additional information do you need to determine \(\Delta E\) for the process? (b) Which quantity is larger for this reaction? (c) Explain your answer to part (b).

The heat of combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l),\) is \(-1367 \mathrm{kJ} / \mathrm{mol} .\) A batch of Sauvignon Blanc wine contains 10.6\(\%\) ethanol by mass. Assuming the density of the wine to be \(1.0 \mathrm{g} / \mathrm{mL},\) what is the caloric content due to the alcohol (ethanol) in a 6 -oz glass of wine \((177 \mathrm{mL})\) ?

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