1) The equilibrium constant, K, for the following reaction is 1.29×10 -2 at 600
ID: 692831 • Letter: 1
Question
1) The equilibrium constant, K, for the following reaction is 1.29×10-2 at 600 K.
COCl2(g) <--- ---> CO(g) + Cl2(g)
An equilibrium mixture of the three gases in a 1.00 L flask at 600 K contains 0.206 M COCl2, 5.16×10-2 M CO and 5.16×10-2M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.88×10-2 mol of CO(g) is added to the flask?
2) The equilibrium constant, K, for the following reaction is 1.80×10-4 at 298 K.
NH4HS(s) <--- ---> NH3(g) + H2S(g)
An equilibrium mixture of the solid and the two gases in a 1.00 L flask at 298 K contains 0.315 mol NH4HS, 1.34×10-2 M NH3 and 1.34×10-2 M H2S. If the concentration of H2S(g) is suddenly increased to 2.12×10-2 M, what will be the concentrations of the two gases once equilibrium has been reestablished?
Explanation / Answer
COCl2(g) <--- ---> CO(g) + Cl2(g)
Q = [CO][Cl2]/[cocl2]
= (0.0516*0.0516)/0.206
= 0.01292
Q = K,system is at equilibrium.
COCl2(g) <--- ---> CO(g) + Cl2(g)
initial 0.206 M 0.0516+0.0288 M 0.0516 M
change +x - x -x
equil 0.206+x M 0.0804-x 0.0516-x
K = ((0.0804-x)*(0.0516-x))/(0.206+x) = 0.0129
x = 0.01115
[COCl2] = 0.206+0.01115 = 0.21715 M
[CO] = 0.0804-x = 0.0804-0.01115 = 0.06925 M
[Cl2] = 0.0516-0.01115 = 0.04045 M
2)
NH4HS(s) <--- ---> NH3(g) + H2S(g)
initial 0.315 M 0.0134 M 0.0134 M
after 0.315 M 0.0134 M 0.0212 M
change +x -x -x
equilb 0.315+x M 0.0134-x 0.0212-x M
K = [NH3][H2S]
1.8*10^-4 = (0.0134-x)*(0.0212-x)
x = 0.00333 M
[NH3] = 0.0134-x = 0.0134-0.00333 = 0.01 M
[H2S] = 0.0212-x = 0.0212-0.00333 = 0.01787 M
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