Thermochemistry

Question

5 O₂(g) + 6 C (graphite) ----> 4 CO₂(g) + 2 CO(g) Delta H₁ = -1798 KJ

CaO(s) + H₂O (l) ------> Ca(OH)₂ (aq) Delta H_{2 =} -65 KJ

CaC₂ (s) + 2 H₂O(l) -----> C₂H₂(g) +Ca(OH)₂ (aq) Delta H₃ = -126 KJ

CaO (s) + 3 C(graphite) -----> CaC₂ (s) + CO Delta H₄ = +463 KJ

Calculate Delta H_rxn for:

2C₂H₂ (g) + 5O_{2 ----> 4 CO}2 (g) +2H₂O(l)

B. Some of the gas C₂H₂ in a 5.0 L cylinder at 25.0 C was removed and burned under standard conditions in the presence of plenty of air.The pressure dropped from 19.7 atto 15.0 atm. How much heat was evolved by this reaction?

Answer 1

Hess' Law

2C₂H₂ (g) + 5O₂ ==> 4CO₂ (g) +2H₂O(l) ... TARGET EQUATION

2C2H2 + 2Ca(OH)2 ==> 2CaC2 + 4H2O ... ∆H = +252

2CaO + 2H2O ==> 2Ca(OH)2 ... ∆H = -130

2CaC2 + 2CO ==> 2CaO + 6C ... -926

5O2 + 6C ===> 4CO2 + 2CO ... ∆H = -1798

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2C2H2 + 2H2O + 5O2 ===> 4H2O + 4CO2

2C2H2 + 5O2 ===> 2H2O + 4CO2 ... TARGET EQUATION

∆H_rxn = 252 - 130 -926 -1798 = -2602 kJ

B. The question is a little confusing because it says some gas was removed from the cylinder and burned under standard conditions (which would be 1 atm and 273K) but then it says the the pressure was 197. atm and the temp was 298K. These aren't standard conditions. Furthermore, when the gas was removed, where was it placed, ie into what volume?

Normally, you could use PV = nRT to solve for n (moles) under the 2 conditions of pressure and subtract the 2 values to tell you moles of gas burned. Then use 2602 kJ/2 mole C2H2 x # moles burned to get your answer, but not sure this is what the question is getting at.