Use the data in the appendix and the following thermochemical equation to determ

Question

Use the data in the appendix and the following thermochemical equation to determine the enthalpy of formation of MgCO3(s). (The ?Hf of MgCl2(s) = -641.60 kJ/mol.)

MgCO3(s) + 2 HCl(g) ? MgCl2(s) + CO2(g) + H2O(l)     ?H = -24 kJ

substance Hfo
kJ/mol Gfo
kJ/mol So
J/mol-K Aluminum: Al(s) 0 0 28.3 AlCl3(s) -704.2 -628.8 110.7 Al2O3(s) -1676 -1582 50.92 Barium: Ba(s) 0 0 28.3 Ba2+(aq) -538.36 -560.7 13 BaCO3(s) -1219 -1139 112 Boron: B(s) 0 0 5.86 BF3(g) -1137.0 -1120.3 254.12 Bromine: Br2(l) 0 0 152.23 Br2(g) 30.91 3.13 245.38 Br(g) 111.9 82.40 174.90 Br1-(aq) -120.9 -102.8 80.71 HBr(g) -36 -53.5 198.59 Calcium: CaO(s) -635.1 -603.5 38.2 Ca(OH)2(s) -986.09 CaCO3(s) -1,206.9 -1,128.8 92.9 Carbon: C(s,graphite) 0 0 5.740 C(s,diamond) 1.987 2.900 2.38 C(g) 716.681 671.289 157.987 CO(g) -110.52 -137.15 197.56 CO2(g) -393.51 -394.36 213.6 CO2(aq) -412.9 -386.2 121 CCl4(g) -103 -53.7 309.7 CCl4(l) -135.4 -68.6 216.4 HCN(g) 135 125 201.7 CH4(g) -74.9 -50.6 186.15 CS2(g) 117.4 67.15 237.7 CS2(l) 89.70 65.27 151.3 C2H2(g) 226.7 209.2 200.8 C2H4(g 52.28 68.12 219.5 C2H6(g) -84.67 -32.89 229.5 CH3OH(g) -201.2 -161.9 237.6 CH3OH(l) -237.6 -166.23 126.8 C2H5OH(g) -235.1 -168.6 282.6 C2H5OH(l) -277.7 -174.76 160.7 CH3COOH(l) -484.5 -390 160 CH3CHO(g) -166 -133.7 266 C4H10(g) -124.73 -15.71 310.0 C4H10(l) -147.6 -15.0 231.0 C6H6(g) 82.927 129.66 269.2 C6H6(l) 49.028 124.50 172.8 C8H18(l) -226 Chlorine: Cl2(g) 0 0 223.0 Cl(g) 121.7 105.7 165.2 Cl1-(aq) -167.2 -131.2 56.5 HCl(g) -92.30 -95.31 186.80 HCl(aq) -167.2 -131.2 56.5 Chromium: Cr2O3(s) -1,138 -1,059 81.2 (NH4)2Cr 2O7(s) -1,807 Copper: Cu(s) 0 0 33.15 CuO(s) -157 -128 42.63 CuSO4(s) -771.36 -661.9 110 Cu2S(s) -79.5 -86.2 120.5 Fluorine: F2(g) 0 0 202.7 F(g) 78.99 61.92 158.64 F1-(aq) -332.6 -278.8 -13.8 HF(g) -271 -273 173.67 Hydrogen: H2(g) 0 0 130.57 H(g) 217.97 203.26 114.60 H1+(aq) 0 0 0 H2O(l) -285.83 -237.2 69.91 H2O(g) -241.82 -228.59 188.7 Iodine: I2(s) 0 0 116.14 I2(g) 62.438 19.36 260.6 I(g) 106.84 70.283 180.68 HI(g) 26.5 1.71 206.48 Iron: Fe(s) 0 0 27.3 FeO(s) -271.9 -255.2 60.75 Fe2O3(s) -824.2 -742.2 87.40 Fe3O4(s) -1117.1 -1014.2 146.4 Lead: Pb(s) 0 0 68.85 Pb2+(aq) 1.6 -24.3 21 PbBr2(s) -277.4 -260.7 161 PbCl2(s) -359 -314 136 PbO(s) -219.0 -189.2 66.5 Magnesium Mg(s) 0 0 32.69 MgO(s) -601.7 -569.4 26.94 MgCl2(s) -641.6 -592.1 89.6 Nitrogen: N2(g) 0 0 191.5 N(g) 472.704 455.5 153.19 NH3(g) -46.11 -16.5 192.3 NH3(aq) -80.29 -26.57 111.3 NH4Cl(s) -314.4 -203.0 94.6 NH4NO3(s) -365.5 -184.0 151.1 NO(g) 90.29 86.60 210.65 NO2(g) 33.84 51.84 240.45 N2O4(g) 9.66 98.28 304.3 Oxygen: O2(g) 0 0 205.03 O(g) 249.17 231.75 160.95 O3(g) 143 163 238.82 Phosphorus: P4(s) 0 0 41.08 P(g) 316.4 280.0 163.2 PCl3(g) -288.07 -269.6 311.7 PCl3(l) -319.6 -272.4 217 PH3(g) 5.4 13.4 210.2 P4O10(s) -1640.1 POCl3(g) -542.2 -502.5 325 POCl3(l) -597.0 -520.9 222 H3PO4(aq) -1288.3 -1142.6 158.2 Potassium K(s) 0 0 64.67 KCl(s) -436.7 -409.2 82.59 KClO3(s) -397.7 -296.3 143.1 Silicon: Si(s) 0 0 18.7 Si(g) 368.2 323.9 167.8 SiO2(s) -910.94 -856.67 41.84 SiH4(g) 34 56.9 204.5 Silver: Ag(s) 0 0 42.55 Ag1+(aq) 105.90 77.11 73.93 AgCl(s) -127.0 -109.70 96.11 Ag2O(s) -31.05 -11.20 121.3 Ag2S(s) -32.6 -40.7 144.0 Sodium: Na(s) 0 0 51.45 Na1+(aq) -240.1 -261.9 59.0 NaCl(s) -411.1 -384.0 72.12 Sulfur: S(s, rhombic) 0 0 31.8 SO2(g) -296.83 -300.19 248.1 SO3(g) -395.72 -371.1 256.6 S2(g) 129 79.30 228.2 H2S(g) -20.6 -33.6 205.7 H2SO4(l) -813.989 -690.101 156.90 H2SO4(aq) -909.3 -744.6 20.08 Tin: Sn(s) 0 0 51.2 SnO(s) -286 -257 56.5 SnO2(s) -580.7 -519.7 52.3 SnCl4(l) -545.2 -474.0 259 SnCl4(g) -471.5 -432.2 366 Titanium: Ti(s) 0 0 30.6 Ti(g) 468 422 180.3 TiO2(s) -944.7 -889.5 50.3 TiCl4(l) -804.2 -737.2 252.3 TiCl4(g) -763.2 -726.8 354 Zinc: Zn(s) 0 0 41.6 ZnO(s) -348.0 -318.2 43.6 ZnCl2(s) -415.1 -369.4 111.5

Explanation / Answer

We will use the relation,

delta Horeaction = delta Hfo(products) - delta Hfo(reactants)

Assuming 'x' represents the enthalpy of formation of MgCO3(s) and feeding all the values given above we get,

-24 kJ = [(-641.60) + (-393.51) + (-285.83)] - [(x) + (2 x -92.30)]

Solving for x we get,

x = -1112.34 kJ/mol

thus, the enthalpy of formation for MgCO3(s) is -1112.34 kJ/mol

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