Question

4. (3 points) Consider the following enthalpies of reaction: CaCz(s) + 2H20 (Ca(OH)2(s) + CH2(g) AH,º=-127.9 kJ/mol Cas) + 1/20 (g) - Cao(s) AH,º=-635.1 kJ/mol Cal(s) + H200 - Ca(OH) (9) AH,=-65.2 kJ/mol The standard enthalpies of combustion of graphite and C2H2(g) are -393.51 and -1299.58 kJ/mol, respectively. Calculate the standard enthalpy of formation of CaCz(s) at 25°C. Aside on enthalpy of combustion: Please note that the standard enthalpy of combustion will form H200, not H:09), if the substance includes H. Example: the standard enthalpy of combustion for benzene is the AH associated with the reaction CH.(I) +15/2 0269) — 6C02(s) + 3H2000

Answer 1

using given 1st reaction in reverse,

(1) Ca(OH)2(s) + C2H2(g) = CaC2(s) + 2H2O(l) [127.9]

using given 3rd reaction,

(2) CaO(s) + H2O(l) = Ca(OH)2(s) [–65.2]

using reverse reaction of C2H2,

(3) 3CO2(g) + H2O(l) = C2H2 (s) + 5/2O2(g) [1299.58]

using reaction of combustion of graphite and multiply by 2,

(4) 2C(s) + 2O2(g) = 2CO2(g) [2 x -393.51]

using given 2nd reaction,

(5) Ca(s) + 1/2 O2(g) = CaO(s) [-635.1]

adding all reactions we will get final reaction as -

2C(s) + Ca(s) = CaC2(s)

for above reaction, standard enthalpy of formation will be = 127.9 - 65.2 + 1299.58 + 2*(-393.51) - 635.1 = -59.84 kJ/mol