HESS LAW AND ENTHALPIES

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

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 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

Hess' Law makes use of which principle to calculate the enthalpy change of a reaction?

• The law of conservation of energy

• The law of conservation of matter

• The law that you will always find a lost item in the last place you look for it

• Murphy's law

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

What is standard enthalpy of formation, ΔH_f^o?

• Heat change when 1 mole of gaseous ions is hydrated in water at 25^oC and 1 atm

• Heat change when 1 mole of compound is formed from its elements in their standard states at 25^oC and 1 atm

• Heat change when 1 mole of gaseous atom is formed from its elements at 25^oC and 1 atm

• Heat change when 1 mole of electrons is removed from 1 mole of gaseous atoms at ground state at 25^oC and 1 atm

Use the H_f values in the table below to answer the following question about the following reaction.

CO_2(g) + N_2(g) ⟶ NO_(g) + CO_(g) ΔH = ?

Formula kJ/mol

CO(g) -110

CO2(g) -394

NO(g) 90

a) Calculate Hrxn for the reaction.

• 370kJ/mol

• 374kJ/mol

• 333kJ/mol

• 335KJ/mol

3 C₂H₂(g) → C₆H₆(g)

What is the standard enthalphy change ΔH^o, for the reaction represented above?

(ΔH^o_f of C₂H₂(g) is 230 kJ mol^-1;

(ΔH^o_f of C₆H₆(g) is 83 kJ mol^-1;)

• -607 kJ

• -147 kJ

• -19 kJ

• +19 kJ

• +773 kJ