Chem 120
Worksheet – Thermochemistry (6 points)
1. A 74.8 g sample of copper at 143.2˚C was added to a well insulated vessel containing 165 mL
of glycerol (C3H8O3(l), density = 1.26 g/mL) at 24.8˚C. The final temperature after mixing was
31.1˚C. The specific heat of copper is 0.393 J/g˚C. Calculate the molar heat capacity of glycerol
assuming no heat loss to the insulated vessel.
2. The thermite reaction:
Fe2O3(s) + 2 Al(s)  Al2O3(s) + 2 Fe(s)
is highly exothermic. 0.500 mole of Fe2O3(s) was mixed with 1.00 mole of Al(s) at room
temperature (25˚C) and the reaction was ignited. The liberated heat was retained within the
products, whose specific heat over a broad temperature range is 0.80 J/g˚C. The melting point of
iron is 1530˚C. Show that the quantity of heat liberated is sufficient to raise the temperature of
the products to the melting point of iron (calculate the temperature to which the products would
rise assuming no heat loss).
3. a. You want to determine the heat capacity of a calorimeter so you can determine accurate
∆H˚’s for reactions. Calculate the heat capacity of the calorimeter using the following
reaction, whose ∆H is known and the data given:
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
∆H˚ = - 57.32 kJ
Amounts used: 50.0 mL of 2.00 M HCl and 50.0 mL of 2.00 M NaOH
Initial temp. of both solutions = 16.9˚C
Maximum temp recorded during reaction = 30.4˚C
Density of resulting NaCl solution: 1.04 g/mL
specific heat of 1.00 M NaCl(aq) = 3.93 J/g-K
b. Use the result from part (a) and the following data to determine the experimental ∆H˚rxn for
the reaction between zinc and HCl(aq):
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
Amounts used: 100.0 mL of 1.00 M HCl and 1.3078 g Zn
Initial temp. of HCl solution and Zn = 16.8˚C
Maximum temp recorded during reaction = 24.1˚C
Density of 1.0 M HCl solution: 1.015 g/mL
specific heat of resulting ZnCl2 solution = 3.95 J/g-K
c. Given the following information, what is the error in your experiment?
∆Hf˚ of ZnCl2(aq) = -4.822 x 102 kJ/mol
Answers: 1) 2.3 x 102 J/mol ˚C 2) amount of heat liberated = 425 kJ, possible Tf would be 5.0
x 103, which is greater than 1530˚C 3 a) 16 J/˚C b) -1.6 x 102 or -1.5 x 102 kJ/mol c) ∆Hrxn =
-1.48 x 102 kJ/mole, 8% error or less
Karen Long Spring 2007