1) The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K
ID: 691969 • Letter: 1
Question
1) The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) <----- ------> PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 500 K contains 0.201 M PCl5, 4.91×10-2 M PCl3 and 4.91×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.27×10-2 mol of Cl2(g) is added to the flask? [PCl5] = M [PCl3] = M [Cl2] = M
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 0.276 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 1.88×10-2 M, what will be the concentrations of the two gases once equilibrium has been reestablished? [NH3] = M [H2S] = M
Explanation / Answer
Kp = [PCl3][Cl2]/(PCl5]
1.2*10^-2 = Kp
initially:
[PCl3] = 4.91*10^-2
[Cl2] = 4.91*10^-2
[PCl5] = 0
in equilbirium
[PCl3] = 4.91*10^-2 - x
[Cl2] = 4.91*10^-2 - x
[PCl5] = 0 + x
substitute
1.2*10^-2 = Kp
[PCl3][Cl2]/(PCl5] = 0.012
(0.0491 - x)(0.0491- x)/x = 0.012
0.0491^2 -2*0.0491x + x^2 = 0.012x
x^2 - ( 0.0491*2+0.012)x + 0.0491^2 = 0
x = 0.03
[PCl3] = 4.91*10^-2 - 0.03 = 0.0191
[Cl2] = 4.91*10^-2 - 0.03 = 0.0191
[PCl5] = 0 + x = 0.03
after we add:
3.27*10^-2 mol of CL2 to the flask
[Cl2] = mol/V = 0.0327/1 = 0.0327 M
initially
[PCl3] = 0.0191
[Cl2] = 0.0191+0.0327 = 0.0518
[PCl5] = 0.03
in equilbirium
[PCl3] = 0.0191 -x
[Cl2] = 0.0518 - x
[PCl5] = 0.03 + x
substitute in K
Kp = [PCl3][Cl2]/(PCl5]
0.012 = ( 0.0191 -x)(0.0518 - x) / (0.03+x)
0.012 *0.03 + 0.012 x = 0.0191 *0.0518 - (0.0518 +0.0191)x + x^2
x^2 + (-0.012-0.0518 -0.0191)x + 0.0191 *0.0518-0.012 *0.03 = 0
X = 0.00845
[PCl3] = 0.0191 -0.00845 = 0.01065
[Cl2] = 0.0518 - 0.00845 = 0.04335
[PCl5] = 0.03 + 0.00845= 0.03845
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