### Thermochemistry Problems:Three Equations Needed

Example #1: A 36.0 g sample of water is initially at 10.0 °C. How much energy is required to turn it into steam at 200.0 °C? (This example starts with a temperature change, then a phase change followed by another temperature change.)

Solution:

q = (36.0 g) (90.0 °C) (4.184 J g¯1 °C¯1) = 13,556 J = 13.556 kJ

q = (40.7 kJ/mol) (36.0 g / 18.0 g/mol) = 81.4 kJ

q = (36.0 g) (100.0 °C) (2.02 J g¯1 °C¯1) = 7272 J = 7.272 kJ

q = 102 kJ (rounded to the appropriate number of significant figures)

Example #2: A 72.0 g sample of ice is at 0 °C. How much energy is required to convert it to steam at 100.0 °C? (This example begins with phase change, then a temperature change and then a second phase change, areas two, three and four on the time-temperature graph.)

Solution:

q = (6.02 kJ/mol) (72.0 g / 18.0 g/mol) = 24.08 kJ

q = (72.0 g) (100.0 °C) (4.184 J g¯1 °C¯1) = 30,125 J = 30.125 kJ

q = (40.7 kJ/mol) (72.0 g / 18.0 g/mol) = 162.8 kJ

q = 217 kJ (rounded to the appropriate number of significant figures)

Example #3: Calculate the heat released by cooling 54.0 g H2O from 57.0 °C to minus 3.0 °C. (Cools from 57.0 to zero, a phase change, the cools from zero to -3; three equations needed.)

Solution:

q = (54.0 g) (57.0 °C) (4.184 J g¯1 °C¯1) = 12,878 J = 12.878 kJ

q = (6.02 kJ/mol) (54.0 g / 18.0 g/mol) = 18.06 kJ

q = (54.0 g) (3.0 °C) (2.06 J g¯1 °C¯1) = 333.73 J = 0.334 kJ

q = 31.3 kJ (rounded to the appropriate number of significant figures)

Example #4: How much energy is required to heat 125.0 g of water from -15.0 °C to 35.0 °C?

The solution:

q = (125.0 g) (15.0 °C) (2.06 J g¯1 °C¯1) = 3862.5 J = 3.8625 kJ

q = (6.02 kJ/mol) (125.0 g / 18.0 g/mol) = 41.806 kJ

q = (125.0 g) (35.0 °C) (4.184 J g¯1 °C¯1) = 18,305 J = 18.305 kJ

q = 64.0 kJ (rounded to the appropriate number of significant figures)

Example #5: How much energy in kJ is needed to heat 5.00 g of ice from -10.0 °C to 30.0 °C? The heat of fusion of water is 6.02 kJ/mol, and the molar heat capacity is 36.6 J/mol K for ice and 75.3 J/mol K for liquid water

Solution:

1) The ice will do three things:

1) heat up from -10.0 °C to 0 °C
2) melt at 0 °C
3) warm up (as a liquid) from 0 to 30.0 °C

2) Each one of those steps requires a calculation. Then, the results of the three calcs will be added together for the final answer.

q1 = (5.00 g / 18.0 g /mol) (10 °C) (36.6 J/mol K)
q2 = (5.00 g / 18.0 g /mol) (6.02 kJ/mol)
q3 = (5.00 g / 18.0 g /mol) (30 °C) (75.3 J/mol K)

q1 and q3 will give J for an answer while q2 gives kJ. Convert q1 and q3 to kJ before adding.

The Celsius and the Kelvin will cancel in the q1 and q3 equations. This is because the size of one °C equals the size of 1 K. The Celsius values in q3 and qc are differences, not specific temperatures. You can confirm this by converting -10 and 0 to their Kelvin values and then subtracting. You will get 10 K.

0.10167 kJ + 1.67 kJ + 0.62750 kJ = 2.40 kJ (to three sig figs)

Example #6: A 10.35 kg block of ice has a temperature of -22.3 °C. The block absorbs 4.696 x 106 J of heat. What is the final temperature of the liquid water?

Solution:

1) Raise the temperature of the ice to 0 °C:

q = (10350 g) (22.3 °C) (2.06 J g¯1 °C¯1)

q = 475458.3 J

2) Melt the ice at 0 °C:

q = (6020 J/mol) (10350 g / 18.015 g/mol)

q = 3458617.818 J

3) How many Joules remain?

total used to this point = 3.934 x 106 J

total remaining: 4.696 x 106 J minus 3.934 x 106 J = 7.62 x 105 J

4) Determine temperature increase of liquid water:

7.62 x 105 J = (10350 g) (x) (4.184 J g¯1 °C¯1)

x = 1.76 °C

Example #7: How many grams of ice at -5.6 °C can be completely converted to liquid at 10.1 °C, if the available heat for this process is 4.74 x 103 kJ?

Solution:

1) Raise the temperature of the ice to zero Celsius:

q = (x) (5.6 °C) (2.06 J g¯1 °C¯1)

2) Melt the ice at zero Celsius:

q = (6020 J/mol) (x / 18.015 g/mol)

3) Raise the temperature of the ice to 10.1 Celsius:

q = (x) (10.1 °C) (4.184 J g¯1 °C¯1)

4) 4.74 x 106 J are used in total:

4.74 x 106 J = [(x) (5.6 °C) (2.06 J g¯1 °C¯1)] + [(6020 J/mol) (x / 18.015 g/mol)] + [(x) (10.1 °C) (4.184 J g¯1 °C¯1)]

The solution for x is left to the reader.

Example #8: Bromine melts at -7.25 °C and boils at 58.8 °C. The enthalpy of fusion of bromine is 10.57 kJ/mol and the enthalpy of vaporization of bromine is 29.96 kJ/mol. The specific heat of liquid bromine is 0.474 J/gK. How much heat, in kJ, is required to convert 25.0 g of solid bromine at -7.25 °C to the gas phase at 58.8 °C?

Solution:

1) Three calculations are needed:

1) melt bromine at -7.25
2) heat liquid bromine from -7.25 to 58.8
3) boil bromine at 58.8

2) Here are the calculation set-ups:

q1 = (25.0 g / 159.808 g/mol) (10.57 kJ/mol)
q2 = (25.0 g) (66.05 K) (0.474 J/gK)
q3 = (25.0 g / 159.808 g/mol) (29.96 kJ/mol)

159.808 g/mol is the molar mass of Br2.
66.05 K is the absolute temperature difference between -7.25 °C and 58.8 °C.
Note that I used it in Kelvin, rather than Celsius.

3) Convert the q2 answer to kJ (from J) and then add all three answers. Round off as appropriate and you're done.