Chemistry: Thermochemical Equations and Calorimeter Data
Using heat, temperature change, and balanced equations
Chemistry: Thermochemical Equations and Calorimeter Data
Using heat, temperature change, and balanced equations
Chemistry - Grade 9-12
- 1
A coffee-cup calorimeter contains 150.0 g of water. A reaction causes the water temperature to increase from 22.4°C to 29.8°C. Calculate the heat gained by the water. Use c = 4.184 J/g°C.
Use q = mcΔT and remember that a temperature increase gives a positive q for the water.
The temperature change is 7.4°C. The heat gained by the water is q = mcΔT = (150.0 g)(4.184 J/g°C)(7.4°C) = 4644 J, or about 4.64 kJ. - 2
In a calorimeter, the water absorbs 3.25 kJ of heat during a reaction. What is the heat change of the reaction, assuming the calorimeter itself absorbs no heat?
The reaction released the same amount of heat that the water gained. The heat change of the reaction is -3.25 kJ. - 3
The thermochemical equation for methane combustion is CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), ΔH = -890 kJ. Is the reaction exothermic or endothermic? Explain.
Look at the sign of ΔH.
The reaction is exothermic because ΔH is negative. A negative enthalpy change means heat is released to the surroundings. - 4
Using the equation CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), ΔH = -890 kJ, calculate the heat released when 2.50 mol of CH4 reacts completely.
The equation releases 890 kJ for every 1 mol of CH4. For 2.50 mol CH4, the heat change is 2.50 × -890 kJ = -2225 kJ, so 2225 kJ of heat is released. - 5
Using the equation 2H2(g) + O2(g) → 2H2O(l), ΔH = -572 kJ, calculate the enthalpy change when 1.00 mol of H2O(l) forms.
Divide the given ΔH by the coefficient of H2O.
The equation shows -572 kJ for forming 2 mol of H2O(l). For 1.00 mol of H2O(l), the enthalpy change is -572 kJ ÷ 2 = -286 kJ. - 6
A reaction is represented by N2(g) + 3H2(g) → 2NH3(g), ΔH = -92.4 kJ. Write the thermochemical equation for the reverse reaction.
The reverse thermochemical equation is 2NH3(g) → N2(g) + 3H2(g), ΔH = +92.4 kJ. Reversing a reaction changes the sign of ΔH. - 7
A student mixes two solutions in a coffee-cup calorimeter. The solution mass is 100.0 g, the specific heat is assumed to be 4.184 J/g°C, and the temperature changes from 20.0°C to 18.6°C. Calculate q for the solution and state whether the reaction absorbed or released heat.
First calculate q for the solution, then use qreaction = -qsolution.
The temperature change is -1.4°C. The solution heat is q = (100.0 g)(4.184 J/g°C)(-1.4°C) = -586 J, so the solution lost 586 J of heat. The reaction absorbed 586 J of heat and is endothermic. - 8
A 25.0 g metal sample is heated to 95.0°C and placed in 100.0 g of water at 22.0°C. The final temperature is 24.5°C. Calculate the heat gained by the water. Use cwater = 4.184 J/g°C.
The water temperature change is 24.5°C - 22.0°C = 2.5°C. The heat gained by the water is q = (100.0 g)(4.184 J/g°C)(2.5°C) = 1046 J. - 9
Using the data from the previous problem, calculate the specific heat of the metal. Assume the heat lost by the metal equals the heat gained by the water.
Use qmetal = -qwater and solve for c.
The metal lost 1046 J of heat. Its temperature change is 24.5°C - 95.0°C = -70.5°C. Using q = mcΔT, -1046 J = (25.0 g)(c)(-70.5°C), so c = 0.594 J/g°C. - 10
A calorimeter has a heat capacity of 45.0 J/°C. During a reaction, its temperature increases by 6.20°C. How much heat does the calorimeter absorb?
The heat absorbed by the calorimeter is q = CΔT = (45.0 J/°C)(6.20°C) = 279 J. - 11
A reaction warms 80.0 g of solution from 19.5°C to 27.0°C in a calorimeter with heat capacity 30.0 J/°C. Assume the solution has c = 4.184 J/g°C. Calculate the total heat absorbed by the solution and calorimeter.
Add qsolution and qcalorimeter.
The temperature change is 7.5°C. The solution absorbs (80.0 g)(4.184 J/g°C)(7.5°C) = 2510 J, and the calorimeter absorbs (30.0 J/°C)(7.5°C) = 225 J. The total heat absorbed is 2735 J, or 2.74 kJ. - 12
For the reaction A + B → C, a reaction coordinate diagram shows the products at a lower energy than the reactants. What is the sign of ΔH, and is the reaction exothermic or endothermic?
The sign of ΔH is negative because the products have lower energy than the reactants. The reaction is exothermic because energy is released. - 13
The decomposition of calcium carbonate is represented by CaCO3(s) → CaO(s) + CO2(g), ΔH = +178 kJ. How much heat is required to decompose 0.750 mol of CaCO3?
Use the mole ratio between CaCO3 and ΔH.
The equation requires 178 kJ for each 1 mol of CaCO3. For 0.750 mol, the heat required is 0.750 × 178 kJ = 134 kJ. - 14
A student dissolves 5.00 g of NH4NO3 in 100.0 g of water. The water temperature decreases from 23.0°C to 17.5°C. Calculate the heat change of the water. Use c = 4.184 J/g°C.
The temperature change is 17.5°C - 23.0°C = -5.5°C. The heat change of the water is q = (100.0 g)(4.184 J/g°C)(-5.5°C) = -2301 J, or -2.30 kJ. - 15
Using the data from the NH4NO3 dissolving problem, calculate the molar enthalpy of solution in kJ/mol. The molar mass of NH4NO3 is 80.04 g/mol. Assume the heat absorbed by dissolving NH4NO3 equals the heat lost by the water.
Find moles first, then divide the heat of the process by moles.
The dissolving process absorbs +2.30 kJ. The moles of NH4NO3 are 5.00 g ÷ 80.04 g/mol = 0.0625 mol. The molar enthalpy of solution is 2.30 kJ ÷ 0.0625 mol = +36.8 kJ/mol.