Answer:
Answer: 1) 155.65 kJ; 2) -59.0 kJ/mol; 3) a)
Explanation:
Answer on Question # 55533 - Chemistry - General chemistry
Question:
1. Given the data. N2(g) + O2(g) = 2 NO(g) ΔH = +180.7 kJ; 2 NO(g) + O2(g) = 2 NO2(g) ΔH = −113.1
kJ; 2 N2O(g) = 2 N2(g) + O2(g) ΔH = − 163.2 kJ; Use Hess’s Law to calculate ΔH for the reaction
N2O(g) + NO2(g) = 3 NO(g); Show your work.
2. Calcium carbide (CaC2) reacts with water to form acetylene (C2H2) and Ca(OH)2. From the
following enthalpy of reaction data and data in Appendix C in textbook, calculate ΔHf° for CaC2(s):
CaC2(s) + 2 H2O(l) Ca(OH)2(s) + C2H2 (g) ΔH = −127.2kJ
3. Using average bond enthalpies, predict which of the following reactions will be most
exothermic: a) C(g) + 2 F2(g) CF4(g) b) CO(g) +3 F2(g) CF4(g) + OF2(g) c) CO2(g) + 4 F2(g) CF4(g) +
2 OF2(g)
Solution
1)
N2(g) + O2(g) = 2 NO(g) ΔH1 = +180.7 kJ x(+2)
2 NO(g) + O2(g) = 2 NO2(g) ΔH2 = −113.1 kJ x(-1)
2 N2O(g) = 2 N2(g) + O2(g) ΔH3 = − 163.2 kJ x(+1)
2N2O(g) + 2NO2(g) = 6 NO(g) ΔH = (ΔH3 + 2ΔH1 – ΔH2)
N2O(g) + NO2(g) = 3 NO(g) ΔH = (ΔH3 + 2ΔH1 – ΔH2)/2 = 155.65 kJ
2)
CaC2(s) + 2 H2O(l) = Ca(OH)2(s) + C2H2 (g) ΔHrxn = −127.2kJ
ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) - ΔHf°(CaC2);
ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol
ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol
3)
a) C(g) + 2 F2(g) = CF4(g)
b) CO(g) +3 F2(g) = CF4(g) + OF2(g)
c) CO2(g) + 4 F2(g) = CF4(g) + 2 OF2(g)
C-F bond enthalpy 440 kJ/mol
C=O bond enthalpy in carbon dioxide 805 kJ/mol
C=O bond enthalpy in carbon monoxide 1077 kJ/mol
O-F bond enthalpy 184 kJ/mol
F-F bond enthalpy 153 kJ/mol
a) ΔHrxn = 2x153 - 4x440 = -1454 kJ – the most exothermic
b) ΔHrxn = 1077 + 3x152 - 2x184 - 4x440 = -595 kJ
c) ΔHrxn =805x2 + 4x153 - 4x440 – 2x2x184 = -274 kJ