Zadanie 3a, Paper 2, maj 2025, TZ3, Chemia IB HL

(i) $[\ce{N2}] = \dfrac{15.0}{25.0} = 0.600$ AND $[\ce{H2}] = \dfrac{22.5}{25.0} = 0.900$ AND $[\ce{NH3}] = \dfrac{5.25}{25.0} = 0.210$ «mol dm⁻³» ✔

$K = \dfrac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3}$ ✔

«$K = \dfrac{0.210^2}{0.600 \times 0.900^3} =$» 0.101 ✔

(ii) sharp decrease in [H₂] at the moment the volume is increased ✔

gradual increase «curve or straight line» up to the marker for the new equilibrium «the equilibrium shifts towards the side with more moles of gas» ✔

final horizontal line at a concentration below the initial concentration ✔

(iii) «ΔG^⊖ = −RTln K = −8.31 × 773 × ln 0.101 =» 1.47 × 10⁴ / 14700 ✔

J mol⁻¹ ✔

(i) M1: the three concentrations; M2: the equilibrium expression; M3: K = 0.101 (accept 0.100-0.101). Award [3] for the correct final answer; also award [3] for Kp = 2.44 × 10^-9 kPa^-2 obtained from partial pressures. (ii) [1] for each feature: instantaneous drop, gradual rise (line or curve), final plateau below the starting level. (iii) M1: 14700 J mol^-1 (accept 1.47 × 10^4 J mol^-1 or 14.7 kJ mol^-1 with the matching unit); M2: correct units (accept J). With K = 0.150 the answer is 12200 J mol^-1. Award [2] for a correct final answer with units.