Hess's Law practice

The enthalpy change for the reaction, ∆H_r , is equal to 

• ∆H₁ + ∆H₂
• ∆H₁ - ∆H₂
• -∆H₁ - ∆H₂
• -∆H₁ + ∆H₂

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 kJ mol^–1

• +111 kJ mol^–1

• -111 kJ mol^–1

• +677 kJ mol^–1

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

Given the following data:

ΔH^○_f[FeO(s)] = –270kJ mol^–1

ΔH^○_f [Fe₂O₃(s)] = –820 kJ mol^–1

Select 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

C₂H₄(g) + H₂(g) -> C₂H₆(g) ∆H°=-137 kJ mol^-1

Which statement about this information is correct?

• The total energy of the bonds broken in the reactants is greater

  than the total energy of the bonds

  formed in the product

• The bonds broken and the bonds made are of the same strength

• The total energy of the bonds broken in the reactants is less than the total energy of the bonds formed in the product

• No conclusion can be made about the sums of the bond enthalpies in the product compared with the reactants