Help with these chem hw questions! If someone could please show how to go about

Question

Help with these chem hw questions! If someone could please show how to go about solving these instead of just listing the answers it'd be much appreciated!

The equilibrium constant, K, for the following reaction is 10.5 at 350 K 2CH2Cl2(8)4e) + CCl4(g) An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.51x102M CH2C2, 0.178 M CH4 and 0.178 M CCI4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 9.49x102 mol of CCl4(g) is added to the flask? [CH2C12] [CH4]

Explanation / Answer

1)

KC = [ccL4][CH4]/[CH2Cl2]^2

    = (0.178^2/0.0551)

    = 0.575

AT EQUILIBRIUM

[CH4] = 0.178 - X

[CCl4] = (0.178+0.0949) - X

[ch2cL2] = 0.0551+2x

0.575 = ((0.178-x)((0.178+0.0949)-x))/(0.0551+2x)^2

x = 0.07 M

[CH4] = 0.178 - 0.07 = 0.108 M

[CCl4] = (0.178+0.0949) - 0.07 = 0.203 M

[ch2cL2] = 0.0551+2*0.07 = 0.1951 M

2)

AT equilibrium

[COCl2] = (19.4*0.136/10.7)+x

[CO] = (19.4*0.08/10.7)-x

[cL2] = (19.4*0.08/10.7)-x

KC = [co][cL2]/[COCl2]

0.047 = ((19.4*0.08/10.7)-x)^2/((19.4*0.136/10.7)+x)

X = 0.0306 M

[COCl2] = (19.4*0.136/10.7)+0.0306 = 0.277 M

[CO] = (19.4*0.08/10.7)-0.0306 = 0.114 M

[cL2] = (19.4*0.08/10.7)-0.0306 = 0.114 M

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