CHM 122 Thermochemistry Practice Problems 1. What is the specific heat, c, (Jg-1K-1) of a substance if 51 J of heat are required to raise the temperature of 1.7 grams of this substance from 20.5 C to 25.5 C? 2. In a calorimetry experiment, the temperature of 500. mL of water rose from 19.78 C to 23.61 C when 1.0 g of sodium metal reacted according to the following balanced equation (on pp 218-219 of McMurry & Fays 4th ed.).

2 Na (s) + H2O (l) 2 NaOH (aq) + H2 (g) Using the assumption that the calorimeters heat capacity is due only to the 500 grams of water (specific heat = 4.184 J/(g K)), calculate qrxn, and Hrxn (in kJ) for the reaction as written. 3. Consider that the following enthalpy changes are known for the stated reactions 1-3 below. Using Hess Law, calculate the enthalpy change for reaction #4?

1. N2 (g) + 2 H2 (g) N2H4 (l) H1 = +50.6 kJ 2. H2 (g) + O2 (g) H2O2 (l) H2 = -187.8 kJ 3. 2 H2 (g) + O2 (g) 2 H2O (g) H3 = -483.6 kJ 4. N2H4 (l) + 2 H2O2 (l) N2 (g) + 4 H2O (g)

CHM 122 Solutions to Thermochemistry Practice Problems 1. What is the specific heat (c, in Jg-1K-1) of a substance if 51 J of heat are required to raise the temperature of 1.7 grams of this substance from 20.5 C to 25.5 C? Solution: c = q / (m T) = 51 J / (1.7 g x 5.0 K) = 6.0 J/(g K) 2. In a calorimeter experiment, the temperature of 500. mL of water rose from 19.78 C to 23.61 C when 1.0 g of sodium metal was added to it. The reaction is stated on pp 218-219 of McMurry & Fays 4th edition:

2 Na (s) + H2O (l) 2 NaOH (aq) + H2 (g) Using the assumption that the calorimeters heat capacity is due only to the 500 grams of water (specific heat = 4.184 J/(g K)), calculate qrxn, and calculate H (in kJ) for the reaction as written. Solution: qcal = 500. g x 4.184 J/(g K) (3.83 K) = 8012 J and qrxn = - qcal = - 8012 J 1.0 g Na is 0.0435 mol so that H = qrxn/nNa = -8012 J/0.0435 mol = -184.3 kJ/mol Note that the reaction, as written, involves two moles of sodium so that the enthalpy change for this reaction is twice the H = -184.3 kJ value for one mole. Thus, the enthalpy change for the reaction as written is H = -368.6 kJ. 3. Consider that the following enthalpy changes are known for the stated reactions 1-3 below. What is the enthalpy change for reaction #4?

1. N2 (g) + 2 H2 (g) N2H4 (l) H1 = +50.6 kJ 2. H2 (g) + O2 (g) H2O2 (l) H2 = -187.8 kJ 3. 2 H2 (g) + O2 (g) 2 H2O (g) H3 = -483.6 kJ 4. N2H4 (l) + 2 H2O2 (l) N2 (g) + 4 H2O (g) H4 = ????

Solution: In order that hydrazine and hydrogen peroxide appear as reactants in rxn #4,

rxn #1 and rxn #2 must be reversed (signs must be changed for H values). Addiitonally, rxn #2 (and the H value) must be doubled because rxn #4 involves two moles of hydrogen peroxide. Lastly, rxn #4 has 4 moles of H2O (g) on the product side, which dictates that we double rxn #3 (and its H value) in our sum to yield rxn #4. All together, then, we have the following sum: (#1 reversed) N2H4 (l) N2 (g) + 2 H2 (g) - H1 = -50.6 kJ (#2 reversed and doubled) 2 H2O2 (l) 2 H2 (g) + 2 O2 (g) -2 H2 = +375.6 kJ (#3 doubled) 4 H2 (g) + 2 O2 (g) 4 H2O (g) 2 H3 = -967.2 kJ sum is reaction #4: N2H4 (l) + 2 H2O2 (l) N2 (g) + 4 H2O (g)

and H4 = - H1 -2 H2 + 2 H3 = -642.2 kJ