Energy cycles in reactions practice questions.

Breaking a chemical bond is a process that is

• exothermic
• endothermic
• neither exothermic nor endothermic
• always spontaneous

Answer: endothermic

Energy must be supplied to break bonds (endothermic); energy is released when new bonds form (exothermic). ΔH depends on the balance of the two.

Given N₂(g) + 2O₂(g) → 2NO₂(g), ΔH = +68 kJ, and N₂(g) + O₂(g) → 2NO(g), ΔH = +180 kJ, use Hess's law to determine ΔH for 2NO(g) + O₂(g) → 2NO₂(g).

• +248 kJ
• -112 kJ
• +112 kJ
• -248 kJ

Answer: -112 kJ

Target = (reaction 1) - (reaction 2): ΔH = (+68) - (+180) = -112 kJ.

Using standard enthalpies of formation (kJ mol⁻¹): CH₄(g) = -75, CO₂(g) = -394, H₂O(l) = -286, determine ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).

• -605 kJ
• -891 kJ
• -755 kJ
• -965 kJ

Answer: -891 kJ

ΔH = ΣΔHf(products) - ΣΔHf(reactants) = [(-394) + 2(-286)] - [(-75) + 0] = -966 - (-75) = -891 kJ.

Given C(s) + O₂(g) → CO₂(g), ΔH = -394 kJ, and H₂(g) + ½O₂(g) → H₂O(l), ΔH = -286 kJ, and CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = -891 kJ, determine ΔHf of methane, C(s) + 2H₂(g) → CH₄(g).

• -75 kJ
• -211 kJ
• +75 kJ
• -1571 kJ

Answer: -75 kJ

ΔHf = ΔHc(C) + 2ΔHc(H₂) - ΔHc(CH₄) = (-394) + 2(-286) - (-891) = -966 + 891 = -75 kJ.

Why can standard enthalpies of formation be used with Hess's law to find the ΔH of any reaction?

• Because formation reactions are always exothermic
• Because enthalpy is a state function, so ΔH is path-independent
• Because bond enthalpies are exact for every compound
• Because all elements have a formation enthalpy of +1 kJ mol⁻¹

Answer: Because enthalpy is a state function, so ΔH is path-independent

Enthalpy is a state function; an alternative route via the elements gives the same overall ΔH, which is the basis of the ΣΔHf(products) - ΣΔHf(reactants) method.

Given ΔHc (kJ mol⁻¹): C(s) = -394, H₂(g) = -286, C₂H₅OH(l) = -1367, determine ΔHf of ethanol, 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l).

• -279 kJ
• -2513 kJ
• +277 kJ
• -687 kJ

Answer: -279 kJ

ΔHf = 2ΔHc(C) + 3ΔHc(H₂) − ΔHc(ethanol) = 2(−394) + 3(−286) − (−1367) = −788 − 858 + 1367 = −279 kJ.