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Ch.9 - Thermochemistry: Chemical Energy
Chapter 9, Problem 11

Calculate the enthalpy change for the reaction C(s) + 2 H2(g) → CH4(g) ΔH = ? Given the enthalpy values for the following reactions CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔH = −890.4 kJ C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ H2(g) + 1/2 O2(g) → H2O (g) ΔH = −285.8 kJ (a) −1569.7 kJ (b) +211.1 kJ (c) −1855.5 kJ (d) −74.7 kJ

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1
Identify the target reaction and the given reactions. The target reaction is C(s) + 2 H2(g) → CH4(g).
Write the given reactions in a way that will help to derive the target reaction. The reactions given are: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), C(s) + O2(g) → CO2(g), and H2(g) + 1/2 O2(g) → H2O(g).
Reverse the first given reaction to form CH4(g) from CO2(g) and H2O(l), as it aligns with the products of the target reaction. This changes the sign of ΔH for this reaction.
Use the second and third given reactions directly as they provide the necessary reactants (C(s) and H2(g)) for the target reaction.
Add the modified first reaction to the second and third reactions, ensuring that all extra reactants and products cancel out, leaving only the desired reactants and products of the target reaction. Sum the ΔH values of these reactions to find the ΔH for the target reaction.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Enthalpy Change (ΔH)

Enthalpy change, denoted as ΔH, represents the heat content change of a system during a chemical reaction at constant pressure. It can be either exothermic (releasing heat, ΔH < 0) or endothermic (absorbing heat, ΔH > 0). Understanding ΔH is crucial for predicting the energy changes associated with reactions, which is essential for calculating the overall enthalpy change in a reaction pathway.
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Hess's Law

Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction, regardless of the pathway taken. This principle allows chemists to calculate the enthalpy change for a reaction that may be difficult to measure directly by using known enthalpy values from related reactions. It is particularly useful in this problem to derive the enthalpy change for the formation of methane from the given reactions.
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Standard Enthalpy of Formation

The standard enthalpy of formation is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. For the reaction in question, the standard enthalpy of formation for CH4(g) is indirectly calculated using the enthalpy values of combustion and formation of reactants and products. This concept is vital for determining the ΔH of the reaction involving carbon and hydrogen to form methane.
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Related Practice
Textbook Question
How much heat is required to raise a 50.0 g piece of iron from 25 °C to its melting point of 1538 °C? The specific heat capacity for iron is 0.451 J/g•°C. (a) 34.1 kJ (b) 168 kJ (c) 12.1 kJ (d) 6.78 kJ
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Textbook Question
A 25.0 g piece of granite at 100.0°C was added to 100.0 g of water of 25.0°C, and the temperature rose to 28.4°C. What is the specific heat capacity of the granite? (The specific heat capacity for water is 4.18 J/(g•°C).) (LO 9.10) (a) 0.563 J/(g•°C) (b) 1.53 J/(g•°C) (c) 0.992 J/(g•°C) (d) 0.794 J/(g•°C)
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Textbook Question

When 12.5 g of NH4NO3 is dissolved in 150.0 g of water of 25.0 °C in a coffee cup calorimeter, the final temperature of the solution of 19.7 °C. Assume that the specific heat of the solution is the same as that of water, 4.18 J/(g•°C). What is the ΔH per mol of NH4NO3? (LO 9.10) NH4NO3 (s) → NH4+ (aq) + NO3−(aq) ΔH = ? (a) +3.60 kJ (b) +23.0 kJ (c) +21.3 kJ (d) −3.60 kJ

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Textbook Question
A table of standard enthalpies of formation (ΔH°f) gives a value of −467.9 kJ/mol for NaNO3(s). Which reaction has a ΔH° value of −467.9 kJ? (a) Na+ (aq) + NO3−(aq) → NaNO3(s) (b) Na(s) + N(g + O3(g) → NaNO3(s) (c) Na(s) + 1/2 N2(g) + 3/2 O2(g) → NaNO3(s) (d) 2 Na(s) + N2(g) + 3 O2(g) → 2 NaNO3(s)
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Textbook Question
What is ΔH for the explosion of nitroglycerin? (LO 9.14) 2 C3H5(NO3)3(l) → 3 N2(g) + 1/2 O2(g) + 6 CO2(g) + 5 H2O(g) (a) −315.0 kJ (b) −4517 kJ (c) −3425 kJ (d) −3062 kJ
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Open Question
The boiling point of a substance is defined as the temperature at which liquid and vapor coexist in equilibrium. Use the heat of vaporization ΔHvap = 30.91 kJ/mol and the entropy of vaporization ΔSvap = 93.2 J/K·mol to calculate the boiling point (°C) of liquid bromine.