CHEY10087 Thermodynamics Workshop and Revision Questions

Answers on Web: http://staff.bath.ac.uk/chsataj

1. A small cylinder of compressed CO 2 weighs 1.04 kg when full and 0.74 kg when empty. If the volume is 250 cm 3, calculate the pressure of the gas at 20 °C

2. The density of a gaseous compound was found to be 1.23 g dm -3 at 330 K and 25.5 kPa. What was the molar mass of the gas?

3. What is the heat capacity of a liquid that rose in temperature from 25 °C to 30.23 °C when supplied with 124 J of heat?

4. Calculate the work done when a gas is compressed from 250 cm 3 to 125 cm 3 by an external pressure of 10 kPa.

5. What is the work done when 10 -3 mol of Ar expands reversibly from a pressure of 150 kPa to a final pressure of 50 kPa at 500 K.

6. A sample of methane of mass 4.5 g occupies 12.7 L at 310 K. Calculate the work done when the gas expands isothermally against a constant external pressure of 200 torr until its volume has increased by 3.3 dm 3. Calculate the work which would be done if the expansion were done reversibly.

7. The heat capacity of air at room temperature (20 °C) is approx. 21 J K -1 mol -1. How much heat is required to raise the temperature of a 5 m x 5 m x 3 m room by 10 °C ? How long will it take a 1 kW heater to achieve this ?

8. A certain liquid has ∆H°(vaporization) = 26 kJ mol -1. Calculate q, w, ∆H and ∆U when 0.50 mol is vaporized at 250K and 750 torr.

9. The standard enthalpy of formation of ethylbenzene is -12 5 kJ mol -1. Calculate its standard enthalpy of combustion. (***)

10. The decomposition at constant volume of 1 mol of KrF 2 gas at 25 °C evolves 59.4 kJ of -1 heat. The enthalpy of sublimation of solid KrF 2 is 41 kJ mol . Calculate ∆H°f, 298 for solid KrF 2.

11. The enthalpy of combustion of gaseous ethane at 298 K is -1558.8 kJ mol -1-. Use mean heat capacities to find the enthalpy of combustion at 373 K. (***).

12. On May 14th 1978, an undergraduate student measured the heat of combustion of -1 naphthalene (C 10 H8) using a bomb calorimeter with a total heat capacity of 10.254 kJ K . The combustion of 647.5 mg of naphthalene with excess oxygen produced a temperature rise of 2.525 K. (a). Calculate the enthalpy change on combustion (b). Calculate the enthalpy of formation of naphthalene

1

13. What will be the sign of the enthalpy change ∆Η and the entropy change, ∆S for the following processes: (a). water freezing to form ice at 0 oC. (b). Water vaporizing to steam at 100 oC (c). The reaction of hydrogen gas and oxygen gas to form water.

-1 14. The standard enthalpy of combustion of phenol (s) is –3054 kJ mol at 298 K. Its standard molar entropy is 144.0 J K -1 mol -1. Calculate the standard Gibbs free energy of formation at 298 K.

Useful data -1 ∆H°f CO 2 = -393.5 kJ mol -1 ∆H°f H 2O = -255.5 kJ mol

-1 -1 Cp C 2H6 = 81.67 J K mol -1 -1 Cp O 2 = 29.355 J K mol -1 -1 Cp CO 2 = 37.11 J K mol -1 -1 Cp H 2O(l) = 75.29 J K mol

2 Answers

-1 1. CO 2 = 44 g mol . Mass = 0.3 kg = 6.82 mol. pV = nRT: p (250 x 10 -6 m3) = (6.82 mol)(8.314 J K -1 mol -1)(293.15 K). p = 6.6 x 10 7 Pa = 66.5 bar

2. Density = 1.23 g dm -3 = 1.23 x 10 3 g m -3 pV = nRT: (n/V) = (RT/p) = (25500 Pa / (8.314 J K -1 mol -1)(122 K)) = 9.294 m 3 mol -1 3 3 -1 1 m = 9.294 mol = 1.23 x 10 g. so M R = 1230 g / 9.294 mol = 132.34 g mol

3. q = c ∆T: c = 124 J / 5.23 K = 23.71 J K -1

4. w = p ∆V = (10000 Pa) (250 – 125) x 10 -6 m 3 = 1.25 J

-3 -1 -1 5. wrev = nRT ln (V fin / V init ) = (10 mol )( 8.314 J K mol )(500 K) ln (3 / 1) = 4.57 J (volume is inversely proportional to pressure)

6. 4.5 g of CH 4 = 4.5 / 16 = 0.2813 mol. w = p ∆V = (200 x 133.32 Pa) (3.3) x 10 -3 m 3 = 87.99 J -1 -1 wrev = nRT ln (V fin / V init ) = (0.2813 mol )( 8.314 J K mol )(310 K) ln (16 / 12.7) = 167.5 J

7. Volume = 5 m x 5 m x 3 m = 75 m 3. From the ideal gas equation (101325 Pa)(75 m 3) = n (8.314 J K -1 mol -1)(293.2 K) n = 3117.5 mol.

-1 -1 Q = n c p ∆T = (3117.5 mol)( 21 J K mol )(10 K) = 654.67 kJ.

1 kW = 1 kJ s -1. Hence heater needs 654.7 s or 10.9 min.

-1 8. ∆H°(vaporization) = q p = 26 kJ mol = 13 kJ for 0.5 mol.

V per mol at these conditions (from Ideal gas equation) = 0.02051m 3. w = p ∆V = (750 torr x 133 32 Pa torr -1) (0.5 mol x 0.02051 m 3 mol -1) = 1025.4 J

∆H = ∆U + ∆n RT . Hence, ∆U = 11.974 kJ.

9. C 6H5C2H5 + 10.5 O 2 → 8 CO 2 + 5 H 2O

∆H°(comb.) = Σ ∆H°f (products) - Σ ∆H°f (reactants)

3 = [ (8 x –393.5) + (5 x –285.5)]-[(-12.5) + (0)] = 4564.5 kJ mol -1

-1 10. (a) KrF 2(g) → Kr (g) + F 2(g) ∆U = -59.4 kJ mol (CONSTANT VOLUME)

-1 (b) KrF2(s) → KrF 2(g) ∆H = +41 kJ mol

From (a), ∆H = ∆U + ∆n RT = -59400 + (1)(8.314)(298) = -56.9 kJ mol -1

∆H°f is the enthalpy change when 1 mol of a compound is formed from its elements in their standard states.

-1 -(a) – (b): Kr (g) + F 2(g) → KrF 2(s) ∆H°f = +56.9 – 41 = 15.9 kJ mol

-1 11. ∆H°298 = -1558.8 kJ mol . ∆H°373 = ∆H°298 + ∆cp (373 – 298) Kirchoff eqn.

-1 -1 C2H6 + 3.5 O 2 → 2 CO 2 + 3 H 2O: ∆cp = 144.76 J K mol

-1 -1 -1 -1 ∆H°373 = -1558800 J mol + (144.76 J K mol ) (373 – 298) K = -1547.9 kJ mol .

12. Bomb calorimeter is at constant volume so that the heat change is ∆U directly.

(a). Energy released = (10.254 kJ K -1) (2.525 K) = -25.892 kJ. This is for 0.6475 g so ∆U = (25.892 kJ) (128.1 g mol -1) / (0.6475 g) = -5122.4 kJ mol -1. [Negative since heat is released]

∆H = ∆U + ∆ ngas RT. C 10 H8 (s) + 12 O 2(g) = 10 CO 2(g) + 4 H 2O(l). Hence, ∆ngas = -2 ∆H = -5122400 J mol -1 + (-2)(8.314 J K -1 mol -1)(298 K) = -5127.4 kJ mol -1.

(b) ∆H°(reaction) = Σ ∆H°f(products) - Σ ∆H°f(reactants)

-5127.4 = [10(-393.5) + 4(-285.8)]-[ ∆H°f (Naphth.) + 0]

-1 ∆H°f (Naphth.) = 49.2 kJ mol .

13. (a). ∆H negative - heat released; ∆ S negative - ice is more ordered than water. (b). ∆H positive - heat needed so added to the water; ∆ S positive - gas is more disordered than liquid. (c). ∆H negative - heat released in an exothermic reaction; ∆ S negative - liquids are more ordered than gases. Also, 3 molecules react to form 2. HOWEVER, this is just for the system. As I showed in lectures, ∆S for the Universe increases. Illustrates the point about being careful to carefully define what we are thinking about.

(b) will be much bigger than (a). It takes more heat to evaporate since molecules have to be moved further apart and gases have greater degrees of molecular motion. The difference in order between a liquid and a gas is much greater than that between a solid and a liquid.

4

14. ∆G° = ∆H° - TS ° - but all need to refer to the formation reaction.

C6H5OH + 7½ O 2 → 6CO 2 + 3H 2O ∆H°= Σ ∆H°f(prod.) - Σ ∆H°f(react)

-3054 = [6(-393.5) + 3(-285.8)]-[ ∆H°f (phenol) + 0] -1 -1 -1 ∆H°f (phenol) = -164.4 kJ mol . S ° = 144.0 J K mol

∆G° = ∆H° - TS ° = (-164.4 x 10 3 J mol -1) – (298.15)( 144.0 J K -1 mol -1) = 207.34

5