Help finding the Hreaction for Enthalpy? - clf 2 - molecular geometry
ClF by oxidation F2 ClF3 important fluorinating agents previously used to produce compounds of uranium in nuclear fuels:
CLF (g) + F2 (g) -----> Cl F3 (l)
Use the following thermochemical equations Hrxn produce calculate ClF3:
ClF 2 (g) + O2 (g) -----> Cl2O (g) + OF2 (g) H = 167.5 kJ
2 F2 (g) + O2 (g) -----> 2 of 2 (g) H = -43.5 kJ
2 Cl F3 (l) + 2 O 2 (g) -----> Cl2O (g) + 2 2 (g) H = 394.1 kJ
Clf 2 - Molecular Geometry Help Finding The Hreaction For Enthalpy?
2 comments:
(A) 2 CLF (g) + O2 (g) -----> Cl2O (g) + OF2 (g) H = 167.5 kJ
(B) 2 F2 (g) + O2 (g) -----> 2 of 2 (g) H = -43.5 kJ
(C) 2 Cl F3 (l) + 2 O 2 (g) -----> Cl2O (g) + 2 2 (g) H = 394.1 kJ
Equation (C) is not balanced, I have the "2 of 2" to "3 OF2"
(A) 2 CLF (g) + O2 (g) -----> Cl2O (g) + OF2 (g) H = 167.5 kJ
(B) 2 F2 (g) + O2 (g) -----> 2 of 2 (g) H = -43.5 kJ
(C) g 2 Cl F3 (l) + 2 O 2 (g) -----> Cl2O (g) + 3 OF2 () H = 394.1 kJ
to receive:
CLF (g) + F2 (g) -----> Cl F3 (l)
1 / 2 the equation (A) is the only way to obtain 1 ClF as a reactant
1 / 2 the equation (B) is the only way to obtain 1 F2 as a reagent
In contrast to 1 / 2 (C) is the only way to obtain 1 as a product ClF3
1 / 2 (A): CLF (g) + 1 / 2 O 2 (g) -----> 1 / 2 Cl2O (g) + 1 / 2 2 (g)
1 / 2 (B) F2 (g) + 1 / 2 O 2 (g) -----> OF2 (g)
- 1 / 2 (C): 1 / 2 Cl2O (g) + 1.5 OF2 (g) -> 1 Cl F3 (L) + 1 O2 (g)
The combination of the two, then everything disappears, except for what I wanted:
CLF (g) + F2 (g) -----> Cl F3 (l)
The Hessw tells us that when the combination of 1 / 2 (A), 1 / 2 (B) -1 / 2 (C) in the form of equations gives us the equation we want to ...
then the combination of that is 1 / 2 (A), DH) 1 / 2 (B, - DH 1 / 2 (C) task will be to us the energy we want:
DH DH rxn = 1 / 2 (A), DH) 1 / 2 (B and - DH 1 / 2 (C)
dH = 1 / 2 (167.5) and 1 / 2 (-43.5) and - 1 / 2 (394.1)
and dH = -21.75 and 83.75 to 197.05
dH = -135.05
sigfigs rounded, your answer is -135.0 kJ
Could you accidentally copied the question wrong? I ask because the equation is not balanced thermochemical past. I think it could be:
2 Cl F3 (l) + 2 O 2 (g) -----> Cl2O 3 OF2
If this is the case, then here's the solution:
Let the equations 1 and 2, as it is and invest 3 equations so ...
ClF 2 (g) + O2 (g) -----> Cl2O (g) + OF2 (g)
2 F2 (g) + O2 (g) -----> 2 of 2 (g)
Cl2O (g) + 3 OF2 (g) -----> 2 Cl F3 (l) + 2 O g) 2 (
Cancel equation on opposite sides of the reaction and the equation that you:
CLF (g) + F2 (g) -----> Cl F3 (l)
Then Hrxn:
Hrxn = H1 + H2 + (H3)
H1 = H-value of the first equation
H2 = H-value of the second equation
H3 = H-value of the third equation
H3 is negative because it reverses the equation to the desired product in the right part of the response given that the value of H is the opposite. So:
2Hrxn = 167.5-43.5 - 394.1 = -270.1 kJ
Thus Hrxn =- 135.05 kJ
If the third equation, as I then copiedYou do not know the answer.
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