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Hess' Law and Enthalpy of Formation Problems

Total questions: 50

Worksheet time: 2hrs 8mins

Name

Class

Date

Using the equations below:

C_(s) + O_2(g) → CO_2(g) ∆H = –390 kJ

Mn_(s) + O_2(g) → MnO_2(s) ∆H = –520 kJ

what is ∆H (in kJ) for the following reaction?

MnO_2(s) + C_(s) → Mn_(s) + CO_2(g)

910

130

-130

-910

Consider the following equations.

Mg_(s) + O_2(g) → MgO_(s) ∆H = –602 kJ

H_2(g) + O_2(g) → H₂O_(g) ∆H = –242 kJ

What is the ∆H value (in kJ) for the following reaction?

MgO_(s) + H_2(g) → Mg_(s) + H₂O_(g)

-844

-360

+360

+844

The enthalpies of combustion of C(s), H₂(g) and C₄H₉OH(l) (in kJ mol^-1) are as follows

C(s) + O₂(g) ⟶ CO₂(g) ∆H = a

H₂(g) + ½O₂(g) ⟶ H₂O(l) ∆H = b

C₄H₉OH(l) + 6O₂(g) ⟶ 4CO₂(g) + 5H₂O(l) ∆H=c

What is the enthalpy change for the reaction shown below?

4C(g) + 5H₂(l) + ½O₂(g) ⟶ C₄H₉OH(l)

c – 4a – 5b

2a + 10b - c

4a + 5b - c

2a + 5b + c

The following equations show the oxidation of carbon and carbon monoxide to carbon dioxide.

C_(s) + O_2(g) → CO_2(g) ΔH = –x kJ mol^–1

CO_(g) + ½ O_2(g) → CO_2(g) ΔH = –y kJ mol^–1

What is the enthalpy change, in kJ/mol, for the oxidation of carbon to carbon monoxide?

2C_(s) + O_2(g) → 2 CO_(g)

x + y

-x - y

2y - 2x

x - y

-2x + 2y

Using the equations below

Cu_(s) + ½ O_2(g) → CuO_(s) ∆H = –156 kJ

2Cu_(s) + 1/2 O_2(g) → Cu₂O_(s) ∆H = –170 kJ

what is the value of ∆H (in kJ) for the following reaction?

(Enter your numerical answer without units)

2CuO_(s) → Cu₂O_(s) + ½O_2(g)

(a)

The standard enthalpy change of formation values of two oxides of phosphorus are:

P_4(s) + 3O_2(g) → P₄O_6(s) ΔH_f = –1600 kJ mol^–1

P_4(s) + 5O_2(g) → P₄O_10(s) ΔH_f = –3000 kJ mol^–1

What is the enthalpy change, in kJ mol^–1, for the reaction below?

P₄O_6(s) + 2O_2(g) → P₄O_10(s)

+4600

+1400

–1400

–4600

The enthalpies of combustion of C(s), H₂(g) and C₄H₉OH(l) (in kJmol^-1) are as follows
C(s) + O₂(g) ⟶ CO₂(g) ∆H=a
H₂(g) + ½O₂(g) ⟶ H₂O(l) ∆H=b
C₄H₉OH(l) + 6O₂(g) ⟶ 4CO₂(g) + 5H₂O(l) ∆H=c
What is the enthalpy change for the reaction shown below?
4C(g) + 5H₂(l) + ½O₂(g) ⟶ C₄H₉OH(l)

c – 4a – 5b

2a + 10b - c

4a + 5b - c

2a + 5b + c

The enthalpy change for the reaction C(s, graphite) + 1⁄2O₂(g) --> CO(g) cannot be measured directly since some carbon dioxide is always formed in the reaction. It can be calculated using Hess’s Law and the enthalpy changes of combustion of graphite and of carbon monoxide. C(s, graphite) + O₂(g) ⟶ CO₂ΔH=-394 kJmol^–1 CO(g) + 1⁄2O₂(g) ⟶ CO₂ΔH=-283 kJmol^–1 The enthalpy change for the reaction of graphite with oxygen to give carbon monoxide is

-677 kJmol^–1

+111 kJmol^–1

-111 kJmol^–1

+677 kJmol^–1

Consider the following equations.

Ca_(s) + O_2(g) → CaO_(s) ∆H = –635 kJ

H_2(g) + O_2(g) → H₂O_(g) ∆H = –242 kJ

What is the ∆H value (in kJ) for the following reaction?

CaO_(s) + H_2(g) → Ca_(s) + H₂O_(g)

-877

-393

+393

+877

10.

The following equations show the oxidation of carbon and carbon monoxide to carbon dioxide.

C_(s) +O_2(g) → CO_2(g) ΔH = –x kJ mol^–1

CO_(g) + O_2(g) → CO_2(g) ΔH = –y kJ mol^–1

What is the enthalpy change, in kJ mol^–1, for the oxidation of carbon to carbon monoxide?

C_(s) + O_2(g) → CO_(g)

x + y

-x - y

y - x

x - y

11.

The data in the image refer to the industrial production of nitric acid from ammonia.

The direct oxidation of ammonia to nitrogen dioxide can be represented by the equation

4NH₃(g) + 7O₂(g) → 4NO₂(g) + 6H₂O(g)

for which the standard enthalpy change, in kJ mol⁻¹, is

−1139

−1024

−794

−679

12.

From the following data,

H₂ (g) + Cl₂ (g) →2HCl(g) ΔH° = -185 kJ,

2H₂ (g) + O₂ (g) →2H₂O(g) ΔH° = -483.7 kJ

calculate ΔH° for the following reaction.

4HCl(g) + O₂ (g) →2Cl₂ (g) + 2H₂O(g)

299 kJ

-114 kJ

-299 kJ

114 kJ

13.

2 NO ⟶ N₂+ O₂ (ΔH = -180.5 kJ)

N₂ + 2 O₂ ⟶ 2 NO₂ (ΔH = + 66.36 kJ)

Calculate the entalphy for :

2 NO + O₂⟶2 NO₂(ΔH = ?)

Is the overall process exothermic or endothermic?

-114.14 kJ,

exothermic

+114.14 kJ, endothermic

246.9 kJ, endothermic

-246.9 kJ, exothermic

14.

From the following enthalpy changes,

XeF₂ (s) → Xe (g) + F₂ (g) ∆H° = +123 kJ

Xe (g) + 2F₂ (g) → XeF₄ (s) ∆H° = -262 kJ

calculate the value of ∆H° for the reaction

XeF₂ (s) + F₂ (g) → XeF₄ (s).

(Enter your numerical answer, without units)

(a)

15.

Given the following data,

N₂ (g) + 3H₂ (g) →2NH₃ (g) ΔH= -92 kJ

2NH₃ (g) + 3Cl₂ (g) →N₂ (g) + 6HCl(g) ΔH = -507 kJ

calculate ΔH° for the following reaction.

6HCl(g) + 3H₂ (g) →3Cl₂ (g) + 6H₂O(g)

415 kJ

-92 kJ

-415 kJ

92 kJ

16.

(Hess's Law) Calculate the ∆H for the following reaction: 2H₂O₂→ 2H₂O + O₂

You are given these two thermochemical equations:

2H₂+ O₂ → 2H₂O ∆H = -572 kJ

H₂ + O₂→ H₂O₂ ∆H = -188 kJ

∆H = -948 kJ

∆H = -196 kJ

∆H = -384 kJ

∆H = -188 kJ

17.

From the data given below

S (s) + O₂(g) ⟶ SO₂ (g) ΔH = -297 kJ mol^-1

SO₂ (g) + ½O₂ (g) ⟶ SO₃ (g) ΔH = -297 kJ mol^-1

Calculate ΔH for the reaction:

S (s) + ³/₂O₂ (g) ⟶ SO₃ (g)

(a)

18.

The enthalpy of reactions for the synthesis two oxides of phosphorus are:

P_4(s) + 3O_2(g) → P₄O_6(s) ΔH_f = –1600 kJ mol^–1

P_4(s) + 5O_2(g) → P₄O_10(s) ΔH_f = –3000 kJ mol^–1

What is the enthalpy change, in kJ mol^–1, for the reaction below?

P₄O_6(s) + 2O_2(g) → P₄O_10(s)

+4600

+1400

–1400

–4600

19.

4 NH₃ (g) + 5 O₂ (g) ⟶ 4 NO (g) + 6 H₂O (g)

Using the following information, calculate ΔH for the reaction shown above:

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

256.0 kJ

-1628.2 kJ

-6387 kJ

None of these

20.

CH₄ (g) + NH₃ (g) ⟶ HCN (g) + 3H₂ (g)

Using the following information, calculate ΔH for the reaction shown above:

N₂(g) + H₂(g) + 2C(s) ⟶ 2 HCN(g) ΔH = +270.3 kJ

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

2 H₂ (g) + C (s) ⟶ CH₄(g) ΔH = -74.9 kJ

256.0 kJ

-1628.2 kJ

-6387 kJ

None of these

21.

2 Al (s) + 3 Cl₂(g) ⟶ 2 AlCl₃ (s)

Using the following information, calculate ΔH for the reaction shown above:

2 Al (s) + 6 HCl (aq) ⟶ 2 AlCl₃(aq) + 3 H₂ (g) ΔH = -1049 kJ

HCl (g) ⟶ HCl (aq) ΔH = -74.8 kJ

H₂ (g) + Cl₂ (g) ⟶ 2 HCl (g) ΔH = -1845 kJ

AlCl₃ (s) ⟶ AlCl₃ (aq) ΔH = -323 kJ

256.0 kJ

-1628.2 kJ

-6387 kJ

None of these

22.

2H₂(g) + 2C (s) + O₂(g) ⟶ C₂H₅OH (l)

Using the following information, calculate ΔH for the reaction shown above:

C₂H₅OH (l) + 2 O₂(g) ⟶ 2 CO₂ (g) + 2 H₂O (l) ΔH = -875 kJ

C (s) + O₂ (g) ⟶ CO₂ (g) ΔH = -394.51 kJ

H₂ (g) + ½ O₂ (g) ⟶ H₂O (l) ΔH = -285.8 kJ

652.0 kJ

-486 kJ

-7687 kJ

None of these

23.

Ca (s) + ½ O₂ (g) + CO₂ (g) ⟶ CaCO₃ (s)

Using the following information, calculate ΔH for the reaction shown above:

Ca (s) + ½ O₂ (g) ⟶ CaO (s) ΔH = -635.1 kJ

CaCO₃(s) ⟶ CaO (s) + CO₂ (g) ΔH = 178.3 kJ

-813.4 kJ

592 kJ

-7687 kJ

None of these

24.

Calculate the ∆H for the following reaction:

2H₂O₂→ 2H₂O + 1 O₂ You are given these two equations: 2H₂+ O₂ → 2H₂O ∆H = -572 kJ H₂ + O₂→ H₂O₂ ∆H = -188 kJ

∆H = -948 kJ

∆H = -196 kJ

∆H = -384 kJ

∆H = -188 kJ

25.

2 NH₃(g) → 3 H₂(g) + N₂(g) ΔH°₂₉₈= 92 kJ/mol_rxn

According to the information above, what is the standard enthalpy of formation, ΔH°_f , for NH₃(g) at 298 K ?

-92 kJ/mol

-46 kJ/mol

46 kJ/mol

92 kJ/mol

184 kJ/mol

26.

The enthalpy change for the reaction 2Al(s)+Fe₂O₃(s)→2Fe(s)+Al₂O₃(s) is −860kJ/mol. Based on the standard enthalpies of formation ΔH°_f provided in the table, what is the approximate ΔH°_f for Fe₂O₃(s) ?

+2540kJ/mol

−2540kJ/mol

+820kJ/mol

−820kJ/mol

27.

C(s)+H₂O(g)→CO(g)+H₂(g)

ΔH°= +131kJ/mol

The reaction between C(s) and H₂O(g) is represented by the balanced chemical equation above. Based on the enthalpy change of the reaction (ΔH°) and the standard heats of formation (ΔH°_f) given in the table below, what is the approximate ΔH°_f for CO(g) ?

−373kJ/mol

−111kJ/mol

+111kJ/mol

+373kJ/mol

28.

Calculate the enthalpy of reaction (ΔH).

4CO_{2 (g)} + 6H₂O _(g) → 2C₂H_{6 (g)} + 7O_{2 (g)}

ΔH_f^o [CO_2(g)] = -393.5 kJ

ΔH_f^o[H₂O_(g)] = -241.8 kJ

ΔH_f^o [C₂H_6(g)] = -84.7 kJ

550.647 KJ/mol

2855.584 KJ/mol

-550.647 KJ/mol

-2855.584 KJ/mol

29.

The standard enthalpy changes of formation of carbon dioxide and of methanoic acid are −394 kJ/mol and −409 kJ/mol respectively. Calculate the enthalpy change for the reaction

H₂ _(g) + CO₂ _(g) → HCOOH _(l)

−803 kJ/mol

−15 kJ/mol

+803 kJ/mol

+15 kJ/mol

30.

When calculating enthalpy change from enthalpy of formation data,

multiply the H_f for each compound by the subscript

multiply the H_f for each compound by the coefficient

31.

Based on the reaction represented by the chemical equation, what is the amount of heat released when 136 g of H₂Sreacts? (Hint: first find enthalpy of formation for reaction, then do stoichiometry)

2 H₂S(g) + 3 O₂(g) ⟶ 2 H₂O(g) + 2 SO₂(g)

2066.2 kJ

4132.4 kJ

2241.8 kJ

4483.6 kJ

32.

Calculate enthalpy (in kJ) for these reactions and determine if reaction is exothermic or endothermic.

2 NO(g) + O₂(g) ⟶ 2 NO₂(g)

114.20, endothermic

114.20, exothermic

-114.20, endothermic

-114.20, exothermic

33.

The enthalpies of formation of C(g), H₂(ℓ) and C₄H₉OH(ℓ) (in kJmol^-1) are as follows

C(g) ∆H_f=a

H₂(ℓ) ∆H_f=b

C₄H₉OH(ℓ) ∆H_f=c

What is the enthalpy change for the reaction shown below?

4C(g) + 5H₂(ℓ) + ½O₂(g) ⟶ C₄H₉OH(ℓ)

c – (4a + 5b)

4a + 5b - c

2a + 10b - c

2a + 5b + c

34.

Calculate the ∆H for the following reaction: 2H₂O₂→ 2H₂O + O₂
You are given these two equations:
2H₂+ O₂ → 2H₂O ∆H = -572 kJ
H₂ + O₂→ H₂O₂ ∆H = -188 kJ

∆H = -948 kJ

∆H = -196 kJ

∆H = -384 kJ

∆H = -188 kJ

35.

Using the equations below:

Zn_(s) + O_2(g) → ZnO_(s) ∆H = –350 kJ

Fe_(s) + O_2(g) → FeO_(s) ∆H = –270 kJ

what is ∆H (in kJ) for the following reaction?

FeO_(s) + Zn_(s) → Fe_(s) + ZnO_(s)

620

-80

-620

36.

Consider the following equations.

Mg_(s) + O_2(g) → MgO_(s) ∆H = –602 kJ

H_2(g) + O_2(g) → H₂O_(g) ∆H = –242 kJ

What is the ∆H value (in kJ) for the following reaction?

MgO_(s) + H_2(g) → Mg_(s) + H₂O_(g)

-844

-360

+360

+844

37.

The following equations show the oxidation of carbon and carbon monoxide to carbon dioxide.

C_(s) +O_2(g) → CO_2(g) ΔH = –x kJ mol^–1

CO_(g) + O_2(g) → CO_2(g) ΔH = –y kJ mol^–1

What is the enthalpy change, in kJ mol^–1, for the oxidation of carbon to carbon monoxide?

C_(s) + O_2(g) → CO_(g)

x + y

-x - y

y - x

x - y

38.

In order to find the enthalpy of combustion of C₃H₈ how must the enthalpy changes be arranged?

ΔH₃ = ΔH₁ + ΔH₂

ΔH₂ = ΔH₃ - ΔH₁

ΔH₁ = ΔH₂ - ΔH₃

0 = ΔH₁ + ΔH₂ + ΔH₃

39.

The standard enthalpy change of formation values of two oxides of phosphorus are:

P_4(s) + 3O_2(g) → P₄O_6(s) ΔH_f = –1600 kJ mol^–1

P_4(s) + 5O_2(g) → P₄O_10(s) ΔH_f = –3000 kJ mol^–1

What is the enthalpy change, in kJ mol^–1, for the reaction below?

P₄O_6(s) + 2O_2(g) → P₄O_10(s)

+4600

+1400

–1400

–4600

40.

Using the equations below

Cu_(s) + ¹/₂O_2(g) → CuO_(s) ∆H = –156 kJ

2Cu_(s) + O_2(g) → Cu₂O_(s) ∆H = –170 kJ

what is the value of ∆H (in kJ) for the following reaction?

2CuO_(s) → Cu₂O_(s) + ¹/₂O_2(g)

142

-15

-142

41.

The standard enthalpy change of formation values of two oxides of phosphorus are:

P_4(s) + 3O_2(g) → P₄O_6(s) ΔH_f = –1600 kJ mol^–1

P_4(s) + 5O_2(g) → P₄O_10(s) ΔH_f = –3000 kJ mol^–1

What is the enthalpy change, in kJ mol^–1, for the reaction below?

P₄O_6(s) + 2O_2(g) → P₄O_10(s)

+4600

+1400

–1400

–4600

42.

CH₄ (g) + NH₃ (g) ⟶ HCN (g) + 3H₂ (g)

Using the following information, calculate ΔH for the reaction shown above:

N₂(g) + H₂(g) + 2C(s) ⟶ 2 HCN(g) ΔH = +270.3 kJ

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

2 H₂ (g) + C (s) ⟶ CH₄(g) ΔH = -74.9 kJ

256.0 kJ

-1628.2 kJ

-6387 kJ

None of these

43.

2 Al (s) + 3 Cl₂(g)⟶ 2 AlCl₃ (s)

Using the following information, calculate ΔH for the reaction shown above:

2 Al (s) + 6 HCl (aq) ⟶ 2 AlCl₃(aq) + 3 H₂ (g) ΔH = -1049 kJ

HCl (g) ⟶ HCl (aq) ΔH = -74.8 kJ

H₂ (g) + Cl₂ (g) ⟶ 2 HCl (g) ΔH = -1845 kJ

AlCl₃ (s) ⟶ AlCl₃ (aq) ΔH = -323 kJ

256.0 kJ

-1628.2 kJ

-6387 kJ

None of these

44.

Ca (s) + ½ O₂ (g) + CO₂ (g) ⟶ CaCO₃ (s)

Using the following information, calculate ΔH for the reaction shown above:

Ca (s) + ½ O₂ (g)⟶ CaO (s) ΔH = -635.1 kJ

CaCO₃(s) ⟶ CaO (s) + CO₂ (g) ΔH = 178.3 kJ

-813.4 kJ

592 kJ

-7687 kJ

None of these

45.

Calculate the enthalpy of reaction (ΔH).

4CO_{2 (g)} + 6H₂O _(g) → 2C₂H_{6 (g)} + 7O_{2 (g)}

ΔH_f^o [CO_2(g)] = -393.5 kJ

ΔH_f^o[H₂O_(g)] = -241.8 kJ

ΔH_f^o [C₂H_6(g)] = -84.7 kJ

550.6 kJ/mol

2855.4 kJ/mol

-550.6 kJ/mol

-3194.2 kJ/mol

46.

Elements in their standard state always have standard enthalpies of ________.

-396 kJ/mol

33.2 kJ/mol

0.0 kJ/mol

396 kJ/mol

47.

The standard enthalpy change for the combustion of graphite is −393.5 kJ/mol and that of diamond is −395.4 kJ/mol.

What is the enthalpy change for the reaction below, in kJ/mol?

C _{(s, graphite)} → C _{(s, diamond)}

−1.9

+1.9

−788.9

+788.9

48.

The standard enthalpy changes of combustion of carbon, hydrogen and methane are shown in the table.
Which one of the following expressions gives the correct value for the standard enthalpy change of formation of methane in kJ mol^–1?
C(s) + 2H₂(g) → CH₄(g)

394 + (2 × 286) – 891

–394 – (2 × 286) + 891

394 + 286 – 891

–394 – 286 + 891

49.

Given the following data: ΔH^○_f[FeO(s)] = –270kJmol^–1
ΔH^○_f [Fe₂O₃(s)] = –820 kJ mol^–1Select the expression which gives the enthalpy change, in kJ mol^–1, for the reaction:
2FeO(s) + 1⁄2O₂(g) → Fe₂O₃(s)

(–820 × 1⁄2) + 270 = –140

(+820 × 1⁄2) – 270 = +140

–820 + (270 × 2) = –280

+820 – (270 × 2) = +280

50.

The enthalpy change for the reaction
C(s, graphite) + 1⁄2O₂(g) ⟶ CO(g) cannot be measured directly since some carbon dioxide is always formed in the reaction.
It can be calculated using Hess’s Law and the enthalpy changes of combustion of graphite and of carbon monoxide.
C(s, graphite) + O₂(g) ⟶ CO₂ΔH=-394 kJmol^–1
CO(g) + 1⁄2O₂(g) ⟶ CO₂ΔH=-283 kJmol^–1
The enthalpy change for the reaction of graphite with oxygen to give carbon monoxide is

-677 kJmol^–1

+111 kJmol^–1

-111 kJmol^–1

+677 kJmol^–1

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Hess' Law and Enthalpy of Formation Problems

Total questions: 50

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