Asked by karen
for the reaction S2F6 (g) 2SF2(g), the equilibrium concentrations are as follows: [S26]=0.000430M, [SF2]=2.08 M, [F2]=1.32M. the equilibrium constant is
Answers
Answered by
DrBob222
You should be more careful in your posts. Half of the information you list either is a typo or incorrect. My answer is based on assumptions as to what you meant to post.No arrow. Equation is missing F2, I assume. S26 I assume is S2F6 and so on into the evening.
Keq = (SF2)^2(F2)/(S2F6)
Keq = (2.08)^2(1.32)/0.000430
Keq = (SF2)^2(F2)/(S2F6)
Keq = (2.08)^2(1.32)/0.000430
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