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

Question

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

An equilibrium mixture of the three gases in a 1.00 L flask at 350 K  contains  5.32E-2 M  CH2Cl2,   0.172 M CH4 and 0.172 M CCl4.   What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.142 mol of CH4(g) is added to the flask?

[CH2Cl2] = M [CH4] = M [CCl4] = M The equilibrium constant, K, for the following reaction is 10.5 at 350 K. An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.32E-2 M CH2Cl2, 0.172 M CH4 and 0.172 M CCl4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.142 mol of CH4(g) is added to the flask?

Explanation / Answer

K=[CCl4][CH4]/[CH2Cl2]^2

or 10.5=(0.172+0.142+x)*(0.172+x)/(5.32*10^-2-2x)^2

or x=-0.008 M

so [CH2Cl2]=0.0692 M

[CH4]=0.306 M

[CCl4]=0.164 M

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