Asked by Veronica
I posted this last night, A 0.500L vessel initially contains 0.0125 mol of H2S2 . Find the equilibrium concentrations of H2 and S2. Kc = 1.67 x 10^-7
2 H2S <--> 2 H2 + S2
I have come up with this equation so far, but I am lost now because I don’t even know if this is right.
1.67x10^-7 = ((4x^2)(x)) / (0.000625-4x^2)
2 H2S <--> 2 H2 + S2
I have come up with this equation so far, but I am lost now because I don’t even know if this is right.
1.67x10^-7 = ((4x^2)(x)) / (0.000625-4x^2)
Answers
Answered by
DrBob222
It's close but not right.
1.67E-7 = (2x)^2(x)/(0.025-2x)^2 is what you started with and that's correct. Your error was in expanding the denominator.
1.67E-7 = 4x^3/(6.25E-4 -0.1x +4x^2)
The next step is to multiply 1.67E-7 x the denominator, then collect the terms in descending order and you end up with a cubic equation. I haven't checked this but I ended up with (and you should go through the math to make sure it's right)
4x^3 -6.68E-7x^2 +1.67E-8x -1/04E-10 = 0
and solve that for x.
1.67E-7 = (2x)^2(x)/(0.025-2x)^2 is what you started with and that's correct. Your error was in expanding the denominator.
1.67E-7 = 4x^3/(6.25E-4 -0.1x +4x^2)
The next step is to multiply 1.67E-7 x the denominator, then collect the terms in descending order and you end up with a cubic equation. I haven't checked this but I ended up with (and you should go through the math to make sure it's right)
4x^3 -6.68E-7x^2 +1.67E-8x -1/04E-10 = 0
and solve that for x.
Answered by
DrBob222
I made a typo in that cubic equation.
4x^3-6.68E-7x^2+1.67E-8x-1.04E-10 = 0
4x^3-6.68E-7x^2+1.67E-8x-1.04E-10 = 0
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