Hess' Law

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 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

For Hess' Law to be used what must be the same for all of the reactions being studied?

• The initial conditions of pressure and temperature.

• The final conditions of pressure and temperature.

• The initial and the final conditions of pressure and temperature.

• The initial and the final conditions of pressure and temperature, and the number of moles of reactants.

Which of the following statements are true for the reaction:

SO_2(g) + ¹/₂O_2(g) ↔ SO_3(g) ΔH = –92 kJ mol^-1

Where ↔ indicates that the reaction can proceed in the forward and the reverse direction.

• The forward and reverse reaction both produce 92 kJ of energy.

• Oxidising 2 moles of SO₂ would produce twice as much energy.

• The reverse reaction has an enthalpy of +92 kJ mol^-1.

• Collecting the SO3 produced in the liquid state would not change the measured enthalpy.

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

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

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

• -15

• -142

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) --> CO₂    ΔH=-394 kJmol^–1
CO(g) + 1⁄2O₂(g) --> CO₂  ΔH=-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 

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