Consider the following reaction: 2H2S (g) + SO2 (g) <----> 3S (s) + 2H2O (g) A r

Question

Consider the following reaction: 2H2S (g) + SO2 (g) <----> 3S (s) + 2H2O (g) A reaction mixture initially containing 0.480 M H2S and 0.480M SO2 was found to contain 1.1*10-3 M H2O at a certain temperature.A second reaction mixture at the same temperature initiallycontains [H2S]=0.245 M and [SO2]=0.330 M. Calculate the equilibrium concentration of H2O in a secondmixture at this temperature. [H2O]=?M
So far I have...
Kc= [H2O]2/[HS]2[SO] = (1.1*10-3)2 / (0.480)3 =1.09*10-5
2H2S (g) + SO2 (g) <----> 3S (s) + 2H2O (g) I: 0.245M 0.330M       0 C: -2x    -x    +2x E: (0.245-2x) (0.330-x)       2x
Kc= (2x)2 / (0.245-2x)2(0.330-x) = 1.09*10-5 4x2 - 1.09*10-3 (0.245-2x)2(0.330-x) = 0
Then I get confused because I don't remember how to solve aquadratic equation with an x3.


2H2S (g) + SO2 (g) <----> 3S (s) + 2H2O (g) A reaction mixture initially containing 0.480 M H2S and 0.480M SO2 was found to contain 1.1*10-3 M H2O at a certain temperature.A second reaction mixture at the same temperature initiallycontains [H2S]=0.245 M and [SO2]=0.330 M. Calculate the equilibrium concentration of H2O in a secondmixture at this temperature. [H2O]=?M
So far I have...
Kc= [H2O]2/[HS]2[SO] = (1.1*10-3)2 / (0.480)3 =1.09*10-5
2H2S (g) + SO2 (g) <----> 3S (s) + 2H2O (g) I: 0.245M 0.330M       0 C: -2x    -x    +2x E: (0.245-2x) (0.330-x)       2x
Kc= (2x)2 / (0.245-2x)2(0.330-x) = 1.09*10-5 4x2 - 1.09*10-3 (0.245-2x)2(0.330-x) = 0
Then I get confused because I don't remember how to solve aquadratic equation with an x3.

Explanation / Answer

We Know that :     2H2S (g) + SO2 (g) <----> 3S (s) +2H2O (g)       Kc = [H2O ]2 / [ H2S ]2 [ SO2 ]              = (1.1*10-3)2 / (0.480)3 =1.09*10-5                     2H2S(g) + SO2 (g) <----> 3S (s) + 2H2O(g) I :      0.245        0.330                           0    C :     -2x         -x                                 +2x      E   0.245-2x       0.330-x                     +2x              1.09*10-5 = (2x)^2 / (0.245-2x)^2 (0.330-x )              solving the above equation for x we get the value of x as               x = 0.000231 M which can be acceptable the other tworoots of the x can be ignored.           [H2O] = 2 x 0.000231 M                       = 0.000462M              

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