Calculate the standard enthalpy of formation of gaseous hydrogen
fluoride (HF) using the following thermochemical information:
C2H4(g) + 6 F2(g) 2 CF4(g) + 4 HF(g) | H = -2486.3 kJ | |
CF4(g) C(s) + 2 F2(g) | H = +680 kJ | |
C2H4(g) 2 C(s) + 2 H2(g) | H = -52.3 kJ |
H = ___?kJ
The answer is not -589.3
The formation reaction is:
1/2 H2(g) + 1/2 F2(g) -> HF(g)
Let this be reaction 0
Lets number the reaction as 0, 1, 2, 3 from top to bottom
required reaction should be written in terms of other reaction
This is Hess Law
required reaction can be written as:
reaction 0 = +0.25 * (reaction 1) +0.5 * (reaction 2) -0.25 * (reaction 3)
So, ΔHo rxn for required reaction will be:
ΔHo rxn = +0.25 * ΔHo rxn(reaction 1) +0.5 * ΔHo rxn(reaction 2) -0.25 * ΔHo rxn(reaction 3)
= +0.25 * (-2486.3) +0.5 * (680.0) -0.25 * (-52.3)
= -268.5 KJ
Answer: -268.5 KJ
Calculate the standard enthalpy of formation of gaseous hydrogen fluoride (HF) using the following thermochemical information:...
Calculate the standard enthalpy of formation of gaseous hydrogen fluoride (HF) using the following thermochemical information: C2H4(g) + 6 F2(g) 2 CF4(g) + 4 HF(g) H = -2486.3 kJ CF4(g) C(s) + 2 F2(g) H = +680 kJ C2H4(g) 2 C(s) + 2 H2(g) H = -52.3 kJ H = ??? kJ
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