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Zadanie 1f

7 pktobliczenioweśrednie (szac.)Paper 2, listopad 2022

Wzorowane na: Chemistry SL Paper 2, November 2022, pytanie 1(f). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.

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Informacja do zadań 1a-1f.

Ammonium chloride, NH4Cl, is a white crystalline solid used as a nitrogen fertilizer for rice crops and as an electrolyte in dry cells.

Polecenie

Solid ammonium chloride decomposes when heated to form two gases, ammonia and hydrogen chloride.

(i) Write the chemical equation for this decomposition.

(ii) Calculate the volume of ammonia produced at STP when an 8.00 g sample of ammonium chloride decomposes. Use sections 2 and 7 of the data booklet.

(iii) Use the thermochemical equations below to calculate the enthalpy change, ΔH⊖, for the decomposition of one mole of solid ammonium chloride.

N2(g) + 3 H2(g) → 2 NH3(g)   ΔH⊖ = −92 kJ

H2(g) + Cl2(g) → 2 HCl(g)   ΔH⊖ = −184 kJ

N2(g) + 4 H2(g) + Cl2(g) → 2 NH4Cl(s)   ΔH⊖ = −628 kJ

(iv) Ammonia and hydrogen chloride recombine on cooling to form solid ammonium chloride. Deduce the Lewis (electron dot) structure of the ammonium ion and state its shape.

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Odpowiedź z klucza

(i) NH4Cl(s) → NH3(g) + HCl(g) ✔

(ii) «n(NHX4Cl)=8.00 g53.50 g mol−1n(\ce{NH4Cl}) = \dfrac{8.00\ \mathrm{g}}{53.50\ \mathrm{g\,mol^{-1}}} =» 0.1495 «mol» «1 : 1 mole ratio» ✔

«V = 0.1495 mol × 22.7 dm3 mol−1 =» 3.39 «dm3» ✔

(iii) «NH4Cl(s) → NH3(g) + HCl(g) is obtained by halving the first two equations and reversing and halving the third»

«ΔH⊖=12(−92)+12(−184)−12(−628)\Delta H^\ominus = \tfrac{1}{2}(-92) + \tfrac{1}{2}(-184) - \tfrac{1}{2}(-628)» ✔

«= −46 − 92 + 314 =» +176 «kJ» ✔

(iv) Lewis structure: nitrogen bonded to four hydrogen atoms by single bonds, no lone pair on nitrogen, in square brackets with a + charge ✔

Shape: tetrahedral ✔

Schemat punktowania

(i) 1 mark; accept the equilibrium arrow. (ii) Award [2] for the correct final answer (accept 3.39–3.41 dm3); accept calculations based on PV = nRT. (iii) Award [2] for the correct final answer. M1: correct treatment of the three equations (halving and reversing as needed); M2: the value +176 kJ. Award [1] for −176 kJ or for +352 kJ (one error in the use of the equations). (iv) M1: Lewis structure with four N–H bonds, no lone pair and the + charge shown; M2: tetrahedral. Allow ECF for the shape only if it fits the Lewis structure drawn.

Komentarz

In (iii) the third equation produces two moles of ammonium chloride, so it has to be halved as well as reversed.

Umiejętności w zadaniu