Solutions to calorimetry problems by RSS. [Text only; no links.]

RSS, June 08, 2006

Calibration problems

1. ΔT = 23.97 °C - 22.45 °C = 1.52 K
C = q / ΔT = 22.5 kJ / 1.52 K = 14.8 kJ/K

2. n = 1.236g / 122.1 g/mol = 0.01023 mol of benzoic acid
q = (-3227 kJ/mol) × (0.01023 mol) = -32.67 kJ
q = - C × ΔT so
C = - q / ΔT = - (-32.67 kJ) / 2.345 K = 13.93 kJ/K

3. n = (2.00 mol/L) × (0.0250 L) = 0.0500 mol
q = (-55.8 kJ/mol) × (0.0500 mol) = - 2.79 kJ
C = - q / ΔT = - (-2.79 kJ) / (11.6 K) = 0.240 kJ/K or 240 J/K

Measuring heats of reaction:

4. ΔT = 23.1 °C - 21.0 °C = 2.1 K
q = - C × ΔT = - (488 J/K) × (2.1 K) = -1025 J or -1.02 kJ
The amount of heat given off was 1.02 kJ.
n = (0.700 mol / L) × (0.0250 L) = 0.0175 mol
ΔH = q / n = (-1.025 kJ) / (0.0175 mol) = -59 kJ for 1 mol of equation

5. q = - C × ΔT = - (240 J/K) × (8.5 K) = -2.04 kJ
(0.100 g) / (24.3 g/mol) = 0.004115 mol
ΔH = q / n = -2.04 kJ / 0.004115 mol = -5.0 x 10² kJ for 1 mol of Mg

6. q = - C × ΔT = - (9.44 kJ/K) × (1.965 K) = -18.55 kJ
n = (0.461 g) / (128.2 g/mol) = 0.003596 mol
ΔH = q / n = (-18.55 kJ) / (0.003596 mol) = -5160 kJ for 1 mol of naphthalene.

© 2006 by Dr. Ronald S. Strange