Medium Heat of Reaction Practice Questions

Concept Explanation

The heat of reaction, also known as enthalpy of reaction (ΔH), is the change in the enthalpy of a chemical system that occurs during a reaction at constant pressure. This value represents the difference between the energy required to break chemical bonds in the reactants and the energy released when new bonds form in the products. If the reaction releases energy, it is considered exothermic and has a negative ΔH; if it absorbs energy, it is endothermic and has a positive ΔH. Understanding this concept is essential for predicting whether a reaction will heat up or cool down its surroundings and for calculating the efficiency of fuels and industrial processes.

To calculate the heat of reaction, chemists often use Hess’s Law, which states that the total enthalpy change for a reaction is the same regardless of the path taken. This allows us to use standard heats of formation (ΔHf°) or bond energy practice questions to determine the overall energy change. For experimental measurements, calorimetry practice questions demonstrate how we use the temperature change of water or a solution to find the heat transferred (q = mcΔT).

Solved Examples

1. Calculating ΔH from Heats of Formation: Calculate the standard heat of reaction for the combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). Given: ΔHf°[CH₄] = -74.8 kJ/mol, ΔHf°[CO₂] = -393.5 kJ/mol, ΔHf°[H₂O(l)] = -285.8 kJ/mol.

   1. Identify the formula: ΔH°rxn = Σ ΔHf°(products) - Σ ΔHf°(reactants).

   2. Calculate products: [1 × (-393.5)] + [2 × (-285.8)] = -393.5 - 571.6 = -965.1 kJ.

   3. Calculate reactants: [1 × (-74.8)] + [2 × 0] = -74.8 kJ (Note: ΔHf° for elements like O₂ is zero).

   4. Subtract: -965.1 - (-74.8) = -890.3 kJ/mol.

2. Calorimetry Calculation: A 5.00 g sample of a metal is heated to 100.0°C and placed in 50.0 g of water at 25.0°C. The final temperature is 28.0°C. Calculate the heat released by the metal. (Specific heat of water = 4.184 J/g°C).

   1. Calculate the heat gained by water: q = m × c × ΔT.

   2. q = 50.0 g × 4.184 J/g°C × (28.0 - 25.0)°C = 50.0 × 4.184 × 3.0 = 627.6 J.

   3. Since heat gained by water equals heat lost by metal: qmetal = -627.6 J.

3. Stoichiometry and Heat: The ΔH for the reaction 2H₂(g) + O₂(g) → 2H₂O(g) is -483.6 kJ. How much heat is released when 10.0 g of H₂ reacts?

   1. Convert mass of H₂ to moles: 10.0 g / 2.02 g/mol = 4.95 mol H₂.

   2. Use the stoichiometric ratio: The reaction shows 2 moles of H₂ release 483.6 kJ.

   3. Set up the calculation: (4.95 mol H₂) × (-483.6 kJ / 2 mol H₂) = -1196.9 kJ.

Practice Questions

1. Calculate the enthalpy change for the reaction: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l). Use ΔHf° values: C₃H₈ = -103.8 kJ/mol, CO₂ = -393.5 kJ/mol, H₂O = -285.8 kJ/mol.

2. In a coffee-cup calorimeter, 100.0 mL of 1.0 M NaOH is mixed with 100.0 mL of 1.0 M HCl. The temperature rises from 22.0°C to 28.7°C. Calculate the ΔH per mole of water formed. (Assume density = 1.0 g/mL and c = 4.18 J/g°C).

3. Carbon disulfide (CS₂) burns in oxygen to form CO₂ and SO₂. If the ΔH for the reaction is -1075 kJ, how many grams of CS₂ must be burned to release 500 kJ of heat?

4. Using the following data: (1) N₂ + O₂ → 2NO (ΔH = +180.5 kJ) and (2) 2NO + O₂ → 2NO₂ (ΔH = -114.1 kJ), find the ΔH for the reaction: N₂ + 2O₂ → 2NO₂.

5. How much energy is required to decompose 45.0 g of CaCO₃ into CaO and CO₂? (ΔH°rxn = +178.3 kJ/mol).

6. When 2.00 g of glucose (C₆H₁₂O₆) is burned in a bomb calorimeter, the temperature of 1.50 kg of water increases by 4.80°C. Calculate the molar heat of combustion of glucose.

7. For the reaction 2Al + Fe₂O₃ → Al₂O₃ + 2Fe, the ΔH is -850 kJ. If 54.0 g of Al reacts with excess Fe₂O₃, what is the total energy released?

8. Calculate the ΔH for the reaction 2C(s) + H₂(g) → C₂H₂(g) given the combustion enthalpies: C(s) = -393.5 kJ, H₂(g) = -285.8 kJ, C₂H₂(g) = -1299.5 kJ.

Answers & Explanations

1. -2220 kJ/mol. ΔH = [3(-393.5) + 4(-285.8)] - [-103.8] = [-1180.5 - 1143.2] + 103.8 = -2220 kJ/mol.

2. -56.1 kJ/mol. Total mass = 200 g. q = 200 × 4.18 × 6.7 = 5601.2 J = 5.60 kJ. Moles HCl/NaOH = 0.100 L × 1.0 M = 0.100 mol. ΔH = -5.60 kJ / 0.100 mol = -56 kJ/mol.

3. 35.4 g. Moles CS₂ = 500 kJ / 1075 kJ/mol = 0.465 mol. Mass = 0.465 mol × 76.14 g/mol = 35.4 g.

4. +66.4 kJ. Add reactions (1) and (2): 180.5 + (-114.1) = +66.4 kJ.

5. 80.2 kJ. Moles CaCO₃ = 45.0 g / 100.09 g/mol = 0.450 mol. Energy = 0.450 mol × 178.3 kJ/mol = 80.2 kJ.

6. -2710 kJ/mol. q = 1500 g × 4.184 J/g°C × 4.80°C = 30124.8 J = 30.12 kJ. Moles glucose = 2.00 g / 180.16 g/mol = 0.0111 mol. ΔH = -30.12 kJ / 0.0111 mol = -2713 kJ/mol.

7. -850 kJ. Moles Al = 54.0 g / 27.0 g/mol = 2.00 mol. Since the reaction uses 2 moles of Al, the total energy released is exactly the ΔH of the reaction: 850 kJ.

8. +226.7 kJ. ΔH = [2(-393.5) + 1(-285.8)] - [-1299.5] = [-787 - 285.8] + 1299.5 = -1072.8 + 1299.5 = +226.7 kJ.

Quick Quiz

Frequently Asked Questions

What is the difference between ΔH and q?

While q represents the heat transferred to or from a system, ΔH (enthalpy change) specifically refers to the heat flow at constant pressure. In most laboratory settings using open containers, these two values are numerically equal.

Why is the heat of formation of elements zero?

By convention, the standard heat of formation for an element in its most stable state at 1 atm and 25°C is defined as zero because no reaction is required to "form" the element from itself. This provides a consistent baseline for standard enthalpy calculations.

How does the physical state of a substance affect the heat of reaction?

The state of matter (solid, liquid, or gas) significantly impacts the enthalpy because phase changes involve energy. For instance, the ΔH of formation for liquid water is different from that of water vapor due to the heat of vaporization.

Can a reaction have a negative ΔH yet feel cold?

No, a reaction with a negative ΔH is exothermic and releases heat into the surroundings, making them feel warmer. If a reaction feels cold, it is endothermic and has a positive ΔH, as it is absorbing thermal energy from its environment.

What is the significance of the "standard" state in ΔH°?

The standard state (indicated by the degree symbol) ensures that measurements are comparable by using a fixed pressure of 1 bar and a specified temperature, usually 298.15 K. This is a fundamental concept discussed in standard chemistry textbooks.

How do catalysts affect the heat of reaction?

Catalysts do not change the heat of reaction (ΔH) because the enthalpy of the reactants and products remains the same. They only lower the activation energy, allowing the reaction to reach equilibrium faster without altering the overall energy balance.

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