Calculating the change in Hrxn? - clf 2- molecular geometry
What is the change in Hrxn for CLF (g) + F2 (g) -> Cl F3 (g) if:
ClF 2 (g) + O2 (g) -> Cl2O (g) + F2O (g) a change in Hrxn = 167.4 kJ / mol
2 Cl F3 (g) + 2 O 2 (g) -> Cl2O (g) + 3 F2O (g) Hrxn = 341.4 kJ / mol
2 F2O (g) - 2 F2 (g) + O2 (g) Hrxn = 43,4 kJ / mol
Monday, January 18, 2010
Clf 2- Molecular Geometry Calculating The Change In Hrxn?
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So, you know how to solve this kind of equation? Lines, such as mathematical formulas and try to remove the matter to try to get the most important equation: CLF + F2 -> Cl F3.
You see, you have a mole on the left of CLF, then the first equation to get two and share it:
ClF + 1/2O2 -> 1/2Cl2O + 1/2F2O as Delta H for this equation is 167.4 / 2, which is 83.7.
You see, you need a mole of F2 to cover the third equation and make it negative delta-H ... and divide by two, since there is only one need:
1/2O2 + 1/2F2 -> F2O as Delta H is negative, according to the equation: -21.7
You can see that you meet the right need ClF3 moles then the second equation and divide by two to one mole of Cl F3:
1/2Cl2O + 3/2F2O -> Cl F3 + O2 delta H = -341.4 / 2 = -170.7
Put together, these three equations to get a larger image.
ClF + 1/2O2 -> 1/2Cl2O + 1/2F2O
1/2O2 + 1/2F2 -> F2O
1/2Cl2O + 3/2F2O -> O2 + Cl F3
Add to these. You see, there are two 1 / 2 O2 and O2 leaving a trace. Total O2 on the right side, so they cancel. They are 1 1 / 2 = 3 / 2 F2O on the right side of the equations and 3 / 2 left ... then disappear. It is 1 / 2 Cl2O both parties to cancel, too ... enter the equation
ClF + F2 -> Cl F3 is what you want! Now ... which must be added the delta-h reaction equations for each handle ... give
83.7 + -21.7 + -170.7 to -108.7kJ/mol!
(Check if my math right ...)
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