How can you determine the heat of transition of Srhombic to Smonoclinic from the following
data using Hess’s law?
i. Srhombic + O2 SO2 ΔH = -291.4 kj
ii. Smonoclinic + O2 SO2 ΔH = -295.4 kj
Answers
Explanation:
Problem #11: Use the standard reaction enthalpies given below to determine ΔH° for the following reaction:
2S(s) + 3O2(g) ---> 2SO3(g)
Given:
SO2(g) ---> S(s) + O2(g) ΔH° = +296.8 kJ
2SO2(g) + O2(g) ---> 2SO3(g) ΔH° = −197.8 kJ
Solution:
1) The two data equations are modified:
S(s) + 2O2(g) ---> 2SO2(g) ΔH° = −593.6 kJ <--- flipped and mult. by 2
2SO2(g) + O2(g) ---> 2SO3(g) ΔH° = −197.8 kJ <--- no change
2) The 2SO2 will cancel out when the equations are added. Add the enthalpies for the answer:
−593.6 + (−197.8) = −791.4 kJ
Problem #12: Calculate ΔH° for:
2C2H4(g) + H2O(ℓ) ---> C4H9OH(ℓ)
Using:
2CO2 + 2H2O(ℓ) ---> C2H4(g) + 3O2(g) ΔH° = +1411.1 kJ
C4H9OH(ℓ) + 6O2(g) ---> 4CO2 + 5H2O(ℓ) ΔH° = −1534.7 kJ
Solution:
1) Both data equations need to be flipped, but let's not flip them yet:
4CO2 + 4H2O(ℓ) ---> 2C2H4(g) + 6O2(g) ΔH° = +2822.2 kJ
C4H9OH(ℓ) + 6O2(g) ---> 4CO2 + 5H2O(ℓ) ΔH° = −1534.7 kJ
I multiplied the first equation by 2.
2) Add the two data equations and their enthalpies:
C4H9OH(ℓ) ---> 2C2H4(g) + H2O(ℓ) ΔH° = +1287.5 kJ
The reaction just above can now be flipped to give us our target equation and the enthalpy of the flipped reaction is −1287.5 kJ.
Problem #13: For the following reaction:
2CO(g) + 2NO(g) ---> 2CO2(g) + N2(g)
Use reactions (a) and (b) to determine ΔH
(a) 2CO(g) + O2(g) ---> 2CO2(g) ΔH = −566.0 kJ
(b) N2(g) + O2(g) ---> 2NO(g) ΔH = 180.6 kJ
Solution:
1) Flip (b) and leave (a) alone:
(a) 2CO(g) + O2(g) ---> 2CO2(g) ΔH = −566.0 kJ
(b) 2NO(g) ---> N2(g) + O2(g) ΔH = −180.6 kJ
2) When you add the two reactions, the O2(g) cancels. Add the two enthalpies for the final answer:
−746.6 kJ
Problem #14: Determine the enthalpy of reaction for the combustion of methane to carbon monoxide:
2CH4(g) + 3O2(g) ---> 2CO(g) + 4H2O(ℓ)
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