Question

# The equilibrium constant, K, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) CH4(g) +...

The equilibrium constant, K, for the following reaction is 10.5 at 350 K.

2CH2Cl2(g) CH4(g) + CCl4(g)

An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.19×10-2 M CH2Cl2, 0.168 M CH4 and 0.168 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.133 mol of CH4(g) is added to the flask?

[CH2Cl2] =____ M

[CH4] = ______M

[CCl4] = ______M

[CCl4] = 0.168 M + 0.133 = 0.301

[CH2Cl2] = 5.19 x 10-2 M = 0.0519 M

_____2 CH2Cl2--->..CH4…+..CCl4

I…0.0519…….......0.168…...0.301..

C…+2x....................-x……….-x

E 0.0519…….......0.168-x…..0.301 –x

Now, K = [CH4][CCl4]/[CH2Cl2]2 = 10.5

=> 10. 5 = 0.168-x *(0.301-x)/(0.0519 + 2x)2

=> 10.5* (0.0519 + 2x)2 = 0.0506 -0.469x + x2

=> 10.5(0.00269 + 0.2076 x + 4x2 )= 0.0506 -0.469x + x2

=> 0.0283 + 2.18 x + 42x2 = 0.0506 -0.469x + x2

=> 41x2 + 2.65 x - 0.0223 = 0

By using quadratic formula only the positive value of x is taken,

=> x = -2.65 + 3.27 / 82

=> x = 0.62 / 82

=> x = 0.00756

Therefore, the concentrations will be--

[CCl4] = 0.301 - 0.00756 = 0.293 M

[CH4] = 0.168 – 0.00756 = 0.160 M

[CH2Cl2] = 0.0519 + 2(0.00756) = 0.0519 + 0.01512 = 0.0670 M

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