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Thermochemistry

Ten questions of mixed difficulty, covering Thermochemistry. Print it, or work through it on screen — the answer key starts on its own page.

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Thermochemistry

Chemistry 30 · maddyhelps.com

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  1. In a calorimeter, 0.0250 mol of a substance reacts and warms 50.0 g of water by 8.00 °C. The molar enthalpy of the reaction is closest to

    1. a) +67.0 kJ/mol
    2. b) −1.68 kJ/mol
    3. c) −67.0 kJ/mol
    4. d) −42.0 kJ/mol

  2. Adding a catalyst to a reaction changes the

    1. a) neither one
    2. b) both the activation energy and ΔH
    3. c) activation energy, but not ΔH
    4. d) ΔH, but not the activation energy

  3. Which of the following is an endothermic process?

    1. a) Burning propane
    2. b) Neutralizing an acid with a base
    3. c) Condensing steam
    4. d) Melting ice

  4. Breaking a chemical bond

    1. a) releases energy
    2. b) neither absorbs nor releases energy
    3. c) releases energy only in exothermic reactions
    4. d) absorbs energy

  5. Which set of units is used for a molar enthalpy of reaction?

    1. a) kJ/mol
    2. b) J/(g·°C)
    3. c) mol/L
    4. d) kJ

  6. On a potential energy diagram for an endothermic reaction, the products are

    1. a) at the same energy as the reactants
    2. b) at the top of the activation energy hump
    3. c) lower in energy than the reactants
    4. d) higher in energy than the reactants

  7. How much heat is absorbed when 100.0 g of water is warmed from 20.0 °C to 30.0 °C? (c = 4.19 J/(g·°C))

    1. a) 0.419 kJ
    2. b) 4.19 kJ
    3. c) 41.9 kJ
    4. d) 12.6 kJ

  8. If the reaction A → B has ΔH = −120 kJ, then the reaction B → A has ΔH =

    1. a) +60 kJ
    2. b) 0 kJ
    3. c) +120 kJ
    4. d) −120 kJ

  9. A 25.0 g sample of metal at 100.0 °C is dropped into 100.0 g of water at 20.0 °C. The final temperature is 22.0 °C. The specific heat capacity of the metal is closest to

    1. a) 0.43 J/(g·°C)
    2. b) 1.68 J/(g·°C)
    3. c) 4.19 J/(g·°C)
    4. d) 0.86 J/(g·°C)

  10. Given C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ; H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ; and CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890.3 kJ, what is ΔH for C(s) + 2H₂(g) → CH₄(g)?

    1. a) −210.9 kJ
    2. b) −74.8 kJ
    3. c) +74.8 kJ
    4. d) −1569.6 kJ

Answer key · Thermochemistry

Chemistry 30 · maddyhelps.com

  1. c) −67.0 kJ/mol — q = (50.0)(4.19)(8.00) = 1676 J = 1.68 kJ absorbed by the water, so the reaction released it: ΔH = −1.68 kJ ÷ 0.0250 mol = −67.0 kJ/mol. The water warming up is what tells you the sign is negative.
  2. c) activation energy, but not ΔH — A catalyst provides a lower-energy pathway, so the hump gets shorter. The reactants and products are unchanged, so the difference between them — ΔH — stays exactly the same.
  3. d) Melting ice — Melting takes energy in to break the attractions holding the solid together. Condensation, combustion and neutralization all release energy.
  4. d) absorbs energy — Bonds are stable arrangements, so pulling one apart always costs energy. Forming bonds releases it, and a reaction is exothermic when the bonds formed release more than the bonds broken absorbed.
  5. a) kJ/mol — Molar enthalpy is energy per mole of a named substance, so kJ/mol. Plain kJ is the energy for the equation as written, and J/(g·°C) is a specific heat capacity.
  6. d) higher in energy than the reactants — Endothermic means energy is absorbed, so it is stored in the products: they sit higher on the diagram. The peak of the hump is the activated complex, not the products.
  7. b) 4.19 kJ — q = mcΔT = (100.0 g)(4.19 J/(g·°C))(10.0 °C) = 4190 J = 4.19 kJ. ΔT is the change in temperature, not the final temperature.
  8. c) +120 kJ — Reversing an equation reverses the sign of its enthalpy change. The energy released going one way has to be supplied to go back.
  9. a) 0.43 J/(g·°C) — Heat lost by the metal equals heat gained by the water: (100.0)(4.19)(2.0) = 838 J. The metal cooled by 78.0 °C, so c = 838 ÷ (25.0 × 78.0) = 0.43 J/(g·°C). Using ΔT = 100 °C for the metal instead of 78 °C is the usual slip.
  10. b) −74.8 kJ — Add the first equation, twice the second, and the reverse of the third: −393.5 + 2(−285.8) + 890.3 = −74.8 kJ. Doubling an equation doubles its ΔH; reversing one flips the sign.