Asked by Margaret
At a particular temperature, 13.7 mol of SO3 is placed into a 3.9-L rigid container, and the SO3 dissociates by the reaction given below.
2 SO3(g) 2 SO2(g) + O2(g)
At equilibrium, 3.8 mol of SO2 is present. Calculate K for this reaction.
2 SO3(g) 2 SO2(g) + O2(g)
At equilibrium, 3.8 mol of SO2 is present. Calculate K for this reaction.
Answers
Answered by
DrBob222
13.7mol/3.9L = about 3.5
3.8/3.9 = about0.97
You need to do these more accurately.
........SO3 ==> 2SO2 + O2
I.......3.5.......0......0
C.......-2x.....2x.......x
E......3.5-2x...2x........x
The problem tells you that at equilibrium (SO2) = 2x = 3.8/3.9 = 0.974M
Knowing x, you can calculate SO2m O2, and SO3, then substitute into the Kc expression and solve for Kc.
3.8/3.9 = about0.97
You need to do these more accurately.
........SO3 ==> 2SO2 + O2
I.......3.5.......0......0
C.......-2x.....2x.......x
E......3.5-2x...2x........x
The problem tells you that at equilibrium (SO2) = 2x = 3.8/3.9 = 0.974M
Knowing x, you can calculate SO2m O2, and SO3, then substitute into the Kc expression and solve for Kc.
Answered by
Margaret
it says my answer is still wrong. I got 0.017683 but it says the right answer is 0.071777081878092.
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