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Ch.9 - Thermochemistry: Chemical Energy
Chapter 9, Problem 10

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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Step 1: First, we need to calculate the heat absorbed or released by the solution using the formula q = mcΔT, where m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature. In this case, the mass of the solution is the mass of the water (150.0 g), the specific heat capacity is 4.18 J/(g•°C), and the change in temperature is the final temperature minus the initial temperature (19.7 °C - 25.0 °C).
Step 2: The heat absorbed or released by the solution is equal to the negative of the heat absorbed or released by the reaction, so q_reaction = -q_solution.
Step 3: To find the ΔH per mol of NH4NO3, we need to divide the heat of the reaction (q_reaction) by the number of moles of NH4NO3. The number of moles of NH4NO3 can be calculated by dividing the mass of NH4NO3 (12.5 g) by its molar mass (80.04 g/mol).
Step 4: The ΔH of the reaction is equal to q_reaction divided by the number of moles of NH4NO3. This will give us the ΔH per mol of NH4NO3.
Step 5: Finally, we need to determine the sign of ΔH. Since the temperature of the solution decreased, the reaction is exothermic and ΔH is negative. If the calculated ΔH per mol of NH4NO3 matches one of the given options, that is the correct answer.

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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 (ΔH) is the heat content change of a system at constant pressure. It indicates whether a reaction is exothermic (releases heat, ΔH < 0) or endothermic (absorbs heat, ΔH > 0). In this context, calculating ΔH for the dissolution of NH4NO3 involves determining the heat absorbed by the solution as the solid dissolves.
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Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. In this scenario, a coffee cup calorimeter is used to measure the temperature change of the solution when NH4NO3 dissolves. The heat absorbed by the solution can be calculated using the formula q = m × c × ΔT, where m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature.
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Molar Mass and Stoichiometry

Molar mass is the mass of one mole of a substance, which is essential for converting grams of a substance to moles. For NH4NO3, the molar mass is approximately 80.04 g/mol. Understanding stoichiometry allows us to relate the amount of heat absorbed (calculated from calorimetry) to the number of moles of NH4NO3 dissolved, enabling the calculation of ΔH per mole.
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Related Practice
Textbook Question

Several processes are given in the table and labeled as endo- thermic or exothermic and given a sign for ∆H°. Which process is labeled with the correct sign of ∆H° and correct classification as endothermic or exothermic? (LO 9.8) Process (d) Rubbing alcohol evaporates from your skin.

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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
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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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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