ΔH°f for O₂(g) at standard state is

Answer: 0 Elements in their standard state have ΔH°f = 0 by definition.

Answer: Σνp ΔH°f (products) − Σνr ΔH°f (reactants) Formation enthalpies: products minus reactants, weighted by stoichiometry.

What is ΔH°rxn for the combustion of methane (using values shown)?

Answer: -890 kJ ΔH°rxn = [(-393) + 2(-286)] - [(-75) + 2(0)] = (-965) - (-75) = -890 kJ.

AP Chemistry · Topic 6.8

Enthalpy of Formation Practice

Part of Thermodynamics.(ENE-3.C)

Practice questions

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Sample questions

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Sample 1difficulty 1/5
$ΔH°_f for O₂(g) at standard state is$
- A
  $Variable$
- B
  $0$
  check_circle
- C
  $+286 kJ/mol$
- D
  $-286 kJ/mol$
Why
$Elements in their standard state have ΔH°_f = 0 by definition.$
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Sample 2difficulty 3/5
$ΔH°_rxn equals$
- A
  $Σν_r ΔH°_f - Σν_p ΔH°_f$
- B
  $ΔS° \times T$
- C
  $Σν_p ΔH°_f (products) - Σν_r ΔH°_f (reactants)$
  check_circle
- D
  $-RT ln K$
Why
$Formation enthalpies: products minus reactants, weighted by stoichiometry.$
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Sample 3difficulty 3/5
$What is ΔH°rxn for the combustion of methane (using values shown)?$
- A
  $-604 kJ$
- B
  $-186 kJ$
- C
  $+890 kJ$
- D
  $-890 kJ$
  check_circle
Why
$ΔH°rxn = [(-393) + 2(-286)] - [(-75) + 2(0)] = (-965) - (-75) = -890 kJ.$
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Sample 4difficulty 3/5
$The diagram shows reactants and products related to elements in their standard states via formation enthalpies.$
$Which formula gives ΔH°rxn from standard enthalpies of formation?$
- A
  $ΣΔHf°(reactants) - ΣΔHf°(products)$
- B
  $ΔHf°(any product alone)$
- C
  $ΣΔHf°(products) + ΣΔHf°(reactants)$
- D
  $ΣΔHf°(products) - ΣΔHf°(reactants)$
  check_circle
Why
$By Hess's law applied to the formation cycle, ΔH°rxn = ΣnΔHf°(products) - ΣmΔHf°(reactants).$
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Sample 5difficulty 4/5
$_f^°$
- A
  $_2$
- B
  $All compounds$
- C
  $_2$
- D
  $_2$
  check_circle
Why
$_f^°$
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