What Was the Initial Quantity of Hydrogen Fluoride in the Reaction Jar?

AI Thread Summary
The equilibrium constant (Keq) for the reaction 2 HF (g) <--> H2 (g) + F2 (g) is given as 4, with 0.045 moles of fluorine present at equilibrium in a 1-liter jar. The initial concentration of hydrogen fluoride (HF) was set as x, leading to the equation 4 = (0.045)(0.045) / (x-0.09)^2. Solving this yields two potential values for x: 0.1125 and 0.0675. However, considering significant figures, the final concentration of HF should be reported as 0.11 M, assuming the volume measurement is exact. The discussion emphasizes the importance of significant figures in reporting the final answer.
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Homework Statement



Keq at a given temperature is 4

2 HF (g) <--> H2 (g) + F2 (g)

A 1 Litre jar contains 0.045 moles of fluorine at equilibrium. What was the initial quantity of hydrogen fluoride in the reaction jar?


2. The attempt at a solution

I set the initial concentration of hydrogen fluoride as x, and filled in the attached ICE table.

Afterwards:

4 = (0.045)(0.045) / (x-0.09)^2

x = 0.1125 and 0.0675

Therefore, the final answer is 0.1125 M
 

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  • ICE Table.png
    ICE Table.png
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Work looks good. Watch your sig figs. 0.090 moles of HF gas disappears, not 0.09 moles. Final answer should also be limited to 2 significant figures. So the final answer is not 0.1125 M but rather 0.11 M. This is assuming that the 1 Liter measurement is an exact measurement (i.e. it doesn't just have one sig fig but is excluded from sig fig considerations in that it is an exact number).
 
Your answer is better than the question itself. It never said you started with HF only. Not that it can be solved without this assumption.
 
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