Unit 8 Study Guide

  1. Energy is the ability to do work or produce heat
  2. Energy cannot be created or destroyedit can only be transformed.
  3. The energy in the universe is constant.
  4. A state function changes independentof the pathway.
  5. Temperature is a measure of the random motion of the

components of a substance.

  1. Heat is measured in joules .
  2. The flow of energy is called heat and it occurs because of a temperature difference
  3. Define the system and the surroundings

The system is the part of the universe we are studying.

The surroundings are everything else in the universe

  1. When heat flows outof the system to its surroundings the process is exothermic.
  2. When heat flows intothe system from its surroundings the process is endothermic.
  3. In any exothermic reaction, some of the potential energy stored in the chemical bonds is converted to thermal energy via heat.
  4. For an endothermic reaction, the potential energy of the reactants would be lower than the potential energy of the products.
  5. First law of thermodynamics is really just the law of conservation of energy
  6. Energy can be changed by the flow of work, heat, or both.
  7. How do we solve for internal energy?

∆E = q + w

  1. What are the two parts of thermodynamic quantities?
  2. a number, giving the magnitude of the change
  3. a sign, indicating the direction of the flow (from the system’s point of view)
  4. –q is exothermic and energy flows to the surroundings.
  5. +q is endothermic and energy flows to the system.
  6. –w is exothermic and work is done by the systemon the surroundings.
  7. +w is endothermic and work is done by the surroundingson the system.
  8. The units of energy are the calorie and the joule.
  9. 1 calorie = 4.184 joules
  10. Specific heat capacityisthe amount of energy required to change the temperature of one gram of a substance by one Celsius degree.
  11. List some substances that have low specific heat capacities.
  12. Convert 15.2 cal to joules
  1. Convert 232 J to calories
  1. A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25.°C to 175°C. Calculate the specific heat capacity of iron.

C = Q = 1086.75 = 0.46 J/g°C

m(Tf-Ti) 15.75(175-25)

  1. Calculate the specific heat capacity of a piece of wood if 1500.0 g of the wood absorbs 67,500 joules of heat, and its temperature changes from 32°C to 57°C.

C = Q = 67500 J = 1.8 J/g°C

m(Tf-Ti) (1500 g)(57-32)

  1. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C?

Q = mC(Tf – Ti) = 10.0g (0.90J/g°C)(55-22) = 297 J

  1. 100.0 g of 4.0°C water is heated until its temperature is 37°C. Calculate the amount of heat energy needed to cause this rise in temperature

Q = mC(Tf-Ti) = 100g(4.184J/g°C)(37 – 4) = 14000 J

  1. What mass of water will change its temperature by 3 0C when 525 J of heat is added to it?

m = Q = 525 J = 40 g

C(Tf–Ti) (4.184J/g°C)(3)

  1. A 0.3 g piece of copper is heated and fashioned into a bracelet. The amount of energy transferred by heat to the copper is 66,300 J. If the specific heat of copper is 390 J/g 0C and the initial temperature of the copper was 15°C what is the final temperature of the copper?

ΔT = Q = 66,300 J = 600°C ΔT = (Tf –Ti) Tf = 615°C

mC 0.3g(390J/g°C) 600°C = (Tf – 15°C)

  1. Enthalpy is used to determine exactly how much energy is transfered.
  2. Theenthalpychange for a reaction under constant pressure is same as the heat (q)for that reaction.
  3. A calorimeter is a device used to determine the heat associated with a chemical reaction.
  4. Hess’s law states that in going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one stepor a series of steps.
  5. The sign of ∆H indicates the direction of heat flow. A exothermic reaction = - ∆H value and an endothermic reaction = + ∆H value
  6. If the reaction is reversed, the sign of ∆H is also reversed.

Use a standard enthalpies of formation table to determine the change in enthalpy for 39 and 40.

Compound / ∆Hf (kJ/mol) / Compound / ∆Hf (kJ/mol)
CO(g) / -110.5 / HCl(g) / -92.3
CO2(g) / -393.5 / H2O(g) / -241.8
NaCl(s) / -411.0 / SO2(g) / -296.1
H2O(l) / -285.8 / NH4Cl(s) / -315.4
  1. NaOH(s) + HCl(g) ----> NaCl(s) + H2O(g)

[(-411.0) + (-241.8)] – [(-426.7) + (-92.3)] = -133.8 kJ

  1. 2 CO(g) + O2(g) ---> 2 CO2(g)

[2(-393.5)] – [2(-110.5)] = -566 kJ

  1. Calculate the enthalpy for the following reaction N2(g)+2O2(g)2NO2 (g), from the following data

N2(g)+O2(g)2NO(g) H = +180kJ

2NO2 (g)2NO(g)+O2(g) H = +112kJ

N2(g)+O2 (g)2NO(g) H = +180kJ

flip 2NO(g)+O2(g)2NO2 (g) H = -112kJ

N2(g)+2O2 (g)2NO2 (g) H= +68kJ

  1. Calculate ΔH for the reaction 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g), from the following data.

N2 (g) + O2 (g) → 2 NO (g) ΔH = -180.5 kJ

N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔH = -91.8 kJ

2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = -483.6 kJ

x2 / 2 N2 (g) + 2 O2 (g) --> 4 NO (g) / H = 2 (-180.5 kJ)
flip x2 / 4 NH3 (g) --> 2 N2 (g) + 6 H2 (g) / H = -2 (-91.8 kJ)
x3 / 6 H2 (g) + 3 O2 (g) --> 6 H2O (g) / H = 3 (-483.6 kJ)
4 NH3 (g) + 5 O2 (g) --> 4 NO (g) + 6 H2O (g) / H = -1628. kJ