Calorimetry Problems – Solutions 1. When 12.29 g of finely divided brass (60% Cu, 40% Zn) at 95.0oC is quickly stirred into 40.00 g of water at 22.0oC in a calorimeter, the water temperature rises to 24.0oC. Find the specific heat of brass. qbrass = -­‐ qsurroundings qsurroundings = qwater (because we are ignoring the heat absorbed by the calorimeter) = 40g x 4.18 J/g°C x 2°C = 334.4 J Therefore, qbrass = -­‐qwater = -­‐334.4 J cbrass = (-­‐334.4J)/(12.29g × (24 – 95)°C) = 0.383 J/g°C 2. In an experiment, 400 mL of 0.600 M HNO3(aq) is mixed with 400 mL of 0.300 M Ba(OH)2(aq) in a constant-­‐pressure calorimeter having a heat capacity of 387 J/oC. The initial temperature of both solutions is the same at 18.88oC, and the final temperature of the mixed solution is 22.49oC. Calculate the heat of neutralization in kJ per mole of HNO3. Acid + Base = neutralization reaction. This question is basically just asking the change in enthalpy of the reaction. qreaction = -­‐ qsurroundings qsurroundings = qsolution + qcalorimeter qsolution = (400g + 400g) x (4.18 J/g°C) x (22.49 – 18.88)°C = 12071.84 J [we assume aqueous solutions to be equivalent to water, so we use density of water and the specific heat of water for our q calculations] qcalorimeter = 1g x 387 J/°C x 3.61°C = 1397.07 J [I think this question is missing some information about the mass of the calorimeter. I assumed it to be 1g. You will be given all the information you need on your tests.] qsurroundings = 12071.84 J + 1397.07 J = 13468.91 J Therefore, qreaction = − 13468.91J = − 13.47 kJ n(HNO3) = c × v = (0.6 M × 0.4 L) = 0.24 mol ΔHneut = (-­‐13.47 kJ)/(0.24 mol) = -­‐ 56.12 kJ/mol 3. When 1.00 dm3 of 1.00 mol dm-­‐3 Ba(NO3)2 solution at 25.0°C is mixed with 1.00 dm3 of 1.00 mol dm-­‐3 Na2SO4 solution at 25.0°C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution is 4.18 J/°C-­‐
g, and that the density of the final solution is 1.0 g/cm3, calculate the enthalpy change per mole of BaSO4 formed. Again, assume solution is equivalent to water. Ignore heat absorbed by the calorimeter. q = 2000g x 4.18 J/g°C x 3.1°C = 25916 J = 25.9 kJ Ba(NO3)2 + Na2SO4 à BaSO4 + 2NaNO3 n(Ba(NO3)2) = n(BaSO4) = c x v = 1 M x 1 L = 1 mol Therefore, ΔH = -­‐25.9 kJ/mol. 4. In a coffee-­‐cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g of water at an initial temperature of 25.00°C. After dissolution of the salt, the final temperature of the calorimeter contents is 23.34°C. Assuming the solution has a heat capacity of 4.18 J/°C-­‐g and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of NH4NO3 in units of kJ/mol. n(NH4NO3) = (1.6g)/(80.05 g/mol) = 0.02 mol mwater = 75g ΔT = (23.34 – 25)°C = -­‐1.66°C à Endothermic!! qsurroundings = 75g x 4.18 J/g°C x (-­‐1.66)°C = -­‐520.41 J = -­‐0.520 kJ qreaction = -­‐ qsurroundings = +0.52 kJ Therefore, ΔH = (+0.52 kJ)/(0.02 mol) = 26 kJ/mol. The value of ΔH is positive because it is an endothermic reaction. Example 1: When a 1.15g sample of anhydrous lithium chloride, LiCl was added to 25.0g of water in a coffee-­‐cup calorimeter, a temperature rise of 3.80K was recorded. Calculate the enthalpy change of solution for 1 mol of lithium chloride. q = 25g x 4.18 J/g K x 3.80 = 397.1 J = 0.397 kJ n(LiCl) = (1.15g)/(42.394 g/mol) = 0.027 mol Therefore, ΔH = (-­‐0.397 kJ)/(0.027 mol) = -­‐14.6 kJ/mol.
Example 2: 180.0 J of heat is transferred to a 100.0g sample of iron, resulting in a temperature rise from 22.0°C to 26.0°C. Calculate the specific heat capacity of iron. ΔT = (26 – 22)°C = 4°C cFe = q/mΔT = (180 J)/(100g x 4°C) = 0.45 J/g°C Practice Problems:
1. 50.0cm3 of 1.00 mol dm-­‐3 hydrochloric acid solution was added to 50.0cm3 of 1.00 mol dm-­‐3 sodium hydroxide solution in a polystyrene beaker. The initial temperature of both solutions was 16.7°C. After stirring and accounting for heat loss, the highest temperature reached was 23.5°C. Calculate the enthalpy change for this reaction. Assume solution to be equivalent to water. ΔT = (23.5 – 16.7)°C = 6.8°C q = (100g x 4.18 J/g°C x 6.8°C) = 2842.4 J = 2.84 kJ n(NaOH) = c x v = (1M x 0.05L) = 0.05 mol ΔH = -­‐2.84 kJ/0.05 mol = -­‐56.8 kJ/mol 2. A student used a simple calorimeter to determine the enthalpy change for the combustion of ethanol. C2H5OH(l) + 3O2 (g) à 2CO2 (g) + 3H2O(l)
When 0.690g (0.015 mol) of ethanol was burned it produced a temperature rise of 13.2K in 250g of water. Calculate ∆𝐻 for the reaction. ΔT = 13.2K q = (250g x 4.18 J/g K x 13.2 K) = 13794 J = 13.79 kJ ΔH = (-­‐13.79 kJ)/(0.015 mol) = -­‐919 kJ/mol 3. 50.0cm3 of 0.200 mol dm-­‐3 copper (II) sulphate solution was placed in a polystyrene cup. After two minutes, 1.20g of powdered zinc was added. The temperature was taken every 30 seconds and the following graph obtained. Calculate the enthalpy change for the reaction taking place. n(Cu2+) = 0.2 M x 0.05 L = 0.01 mol n(Zn) = 1.20 g/65.37 g/mol = 0.0184 mol [Cu2+ is the limiting reagent] ΔT = 11°C (approx..) q = 50g x 4.18 J/g°C x 11°C = 2299 J = 2.299 kJ ΔH = -­‐2.299 kJ/0.01 mol = -­‐229.9 kJ/mol.