The student measures delta H3 directly and gets -54.0 kJ/mol. The discrepancy from the Hess's-law prediction is best attributed to:

Answer: heat loss to surroundings during the most exothermic run causing magnitudes to be slightly underestimated Heat loss to the surroundings systematically lowers measured deltaT and so |delta H|, especially in the more exothermic run (2). The 1.5 kJ discrepancy is consistent with such loss; Hess's law itself is exact.

If ΔH for 2 X → Y is −80 kJ, ΔH for X → ½Y is

Answer: −40 kJ Halving stoichiometry halves ΔH.

Using Hess's law, what is ΔH₁ for A + B → C?

Answer: -20 kJ ΔH₁ = ΔH₂ + ΔH₃ = (-50) + (+30) = -20 kJ.

AP Chemistry · Topic 6.9

Hess's Law Practice

Part of Thermodynamics.(ENE-3.D)

Practice questions

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Sample questions

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Sample 1difficulty 1/5
$A student measures three reactions with a coffee-cup calorimeter to verify Hess's law for the dissolution of NaOH and reaction with HCl: (1) NaOH(s) -> NaOH(aq), delta H1 = -44.5 kJ/mol (2) NaOH(s) + HCl(aq) -> NaCl(aq) + H2O(l), delta H2 = -100.0 kJ/mol (3) NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l), delta H3 = ? Reactions (1) and (2) used 2.00 g of NaOH each; (3) used 50.0 mL of 1.00 M NaOH and 50.0 mL of 1.00 M HCl. Combine results to predict delta H3.$
$The student measures delta H3 directly and gets -54.0 kJ/mol. The discrepancy from the Hess's-law prediction is best attributed to:$
- A
  $phenolphthalein interfering$
- B
  $rounding errors in molar masses only$
- C
  $Hess's law not applying to aqueous reactions$
- D
  $heat loss to surroundings during the most exothermic run causing magnitudes to be slightly underestimated$
  check_circle
Why
$Heat loss to the surroundings systematically lowers measured deltaT and so |delta H|, especially in the more exothermic run (2). The 1.5 kJ discrepancy is consistent with such loss; Hess's law itself is exact.$
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Sample 2difficulty 3/5
$If ΔH for 2 X \to Y is -80 kJ, ΔH for X \to ½Y is$
- A
  $+40 kJ$
- B
  $-40 kJ$
  check_circle
- C
  $-80 kJ$
- D
  $-160 kJ$
Why
$Halving stoichiometry halves ΔH.$
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Sample 3difficulty 3/5
$A + B \to D + E ΔH₂ = -50 kJ D + E \to C ΔH₃ = +30 kJ (reverse: -30)$
$Using Hess's law, what is ΔH₁ for A + B \to C?$
- A
  $+80 kJ$
- B
  $-20 kJ$
  check_circle
- C
  $-80 kJ$
- D
  $+20 kJ$
Why
$ΔH₁ = ΔH₂ + ΔH₃ = (-50) + (+30) = -20 kJ.$
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Sample 4difficulty 3/5
$Given A \to B (ΔH = +20) and B \to C (ΔH = -50), ΔH for A \to C is$
- A
  $-30 kJ$
  check_circle
- B
  $+30 kJ$
- C
  $+70 kJ$
- D
  $-70 kJ$
Why
$Sum: 20 + (-50) = -30 kJ.$
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Sample 5difficulty 3/5
$For CH₄(g) + 2 O₂(g) \to CO₂(g) + 2 H₂O(l), use ΔH_f° values.$
$What is ΔH for the combustion of methane?$
- A
  $-890.3 kJ$
  check_circle
- B
  $-604.5 kJ$
- C
  $+890.3 kJ$
- D
  $-1184 kJ$
Why
$ΔH = [(-393.5) + 2(-285.8)] - [(-74.8) + 0] = (-393.5 - 571.6) + 74.8 = -965.1 + 74.8 = -890.3 kJ.$
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