Zadanie 3a

(i) bonds broken: 4 C−C + 12 C−H + 8 O=O = 4×346 + 12×414 + 8×498 = 10336 «kJ» ✔

bonds formed: 10 C=O + 12 O−H = 10×804 + 12×463 = 13596 «kJ» ✔

ΔH = «10336 − 13596 =» −3260 «kJ mol⁻¹» ✔

(ii) n(C₅H₁₂) = 45.0/72.17 = 0.6235 mol AND n(O₂) = 128/32.00 = 4.00 mol ✔

all the pentane would need 8 × 0.6235 = 4.988 mol of O₂, more than the 4.00 mol available, so O₂ is the limiting reactant ✔

$n(\ce{CO2}) = \tfrac{5}{8} \times 4.00 = 2.50$ mol; m(CO₂) = 2.50 × 44.01 = 110 «g» ✔

(iii) bond enthalpies are average values «taken over many compounds, so the calculated ΔH is only approximate» ✔

OR in (i) water is a gas AND the section 14 value refers to liquid water ✔ «condensation of water vapour is exothermic»

OR the section 14 value refers to liquid pentane AND (i) used gaseous pentane ✔

(i) [3] M1: bonds broken correctly counted and summed; M2: bonds formed correctly counted and summed; M3: ΔH = broken − formed with the correct sign. Award [3] for the correct final answer. Award [2 max] for a final answer of +3260 kJ mol-1 (sign error). (ii) [3] M1: both amounts, of pentane and of oxygen, correct; M2: oxygen identified as the limiting reactant, with the comparison shown; M3: correct mass of CO2. Award [3] for the correct final answer; accept 109.5-110.5 g; ECF throughout. Do not penalize the use of whole-number molar masses. (iii) [1] any one of: bond enthalpies are averages; water as a gas in (i) but as a liquid in section 14 (condensation is exothermic); pentane liquid in section 14 but gas in the equation. Accept 'the enthalpy of combustion data are more accurate/are not average values'.