An equilibrium mixture of the three gases in a 1.00 L flask at 600 K contains 0.

Question

An equilibrium mixture of the three gases in a 1.00 L flask at 600 K contains 0.212 M COCl_2, 5.23 times 10^-2 M CO and 5.23 times 10^-2 M Cl_2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.63 times 10^- mol of Cl_2 (g) is added to the flask? [COCl_2] = M [CO] = M [Cl_2] = M The equilibrium constant, K, for the following reaction is 10.5 at 350 K. 2CH_2 Cl_2 (g) CH_4(g) + CCI_4 (g) An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.31 times 10^-2 M CH_2 CI_2, 0.172 M CH_4 and 0.172 M CCI_4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 8.96 times 10^-2 mol of CH_4(g) is added to the flask? [CH_2 C_I2] = M [CH_4] = M [CCI_4] = M

Explanation / Answer

Question 1.

The reaction:

COCl2 = CO + Cl2

K = [CO ] [Cl2 ]/[COCl2 ]

in equilbrium

K = (5.23*10^-2)(5.23*10^-2)/(0.212) = 0.012902

After we add:

[COCl2 ] = 0.212

[CO ] = 5.23*10^-2

[CL2 ] = 5.23*10^-2

addition of 3.63*10^-2 mol

[COCl2 ] = 0.212+ 3.63*10^-2 = 0.2483 M

[CO ] = 5.23*10^-2

[CL2 ] = 5.23*10^-2

In equilbrium:

[COCl2 ] = 0.2483 -x

[CO ] = 5.23*10^-2 + x

[CL2 ] = 5.23*10^-2 + x

Substitutein Kc:

K = [CO ] [Cl2 ]/[COCl2 ]

0.012902 = (5.23*10^-2 + x)(5.23*10^-2 + x) /(0.2483 -x )

solve for x

0.0523^2 + 2*0.0523x + x^2 = 0.012902*0.2483 - 0.012902x

x^2 + (2*0.0523 + 0.012902)x + 0.0523^2 -0.012902*0.2483 = 0

x^2 + 0.117502x + -0.0004682= 0

x = 0.00385

[COCl2 ] = 0.2483 - 0.00385 = 0.24445 M

[CO ] = 5.23*10^-2 + 0.00385 = 0.05615 M

[CL2 ] = 5.23*10^-2 +0.00385 = 0.05615 M

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