The equilibrium constant, K, for the following reaction is 1.20 times 10^-2 at 5

Question

The equilibrium constant, K, for the following reaction is 1.20 times 10^-2 at 500 K. PCl_5(g) PCl_3(g) + Cl_2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 500 K contains 0.196 M PC1_5, 4.86 times 10^-2 M PC1_3 and 4.86times l0^-2 M Cl_2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.20 times 10^-2 mol of PCl_3(g) is added to the flask? [PCl_5] = M [PCl_3] = M [Cl_2] = The equilibrium constant, K, for the following reaction is I.80 times l0^-4 at 298 K. NH_4 HS(s) NH_3(g) + H_2 S(g) An equilibrium mixture of the solid and the two gases in a 1.00 L flask at 298 K contains 0.296 mol NH4HS, 1.34 times 10^-2 M NH_3 and 1.34 times 10^-2 M H_2 S. If the concentration of NH_3(g) is suddenly increased to 2.32 times l0^-2 M, what will be the concentrations of the two gases once equilibrium has been reestablished? [NH_3] = M [H_2 S] = M The equilibrium constant, K, for the following reaction is 1.80 times 10^-2 at 698 K. 2HI(g) H_2(g) + I_2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.329 M HI, 4.41 times l0^-2 M H_2 and 4.41 times l0^-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.216 mol of HI(g) is added to the flask? [HI] = M [H_2] = M [I_2] = M

Explanation / Answer

                         PCl5(g)   <====> Pcl3(g) + cl2(g)

at equil               0.196 M          0.0486M   0.0486 M

after reestablished       0.196+x          0.0806-x   0.0486-x

kc = 0.0486^2/0.196 = 0.012

0.012 = ((0.0806-x)(0.0486-x))/(0.196+x)

x = 0.012 M

after reestablished

concentration of PCL5 = 0.196 +0.012 = 0.208 M

concentration of PCL3 = 0.0806-0.012 = 0.0686 M

concentration of cl2 = 0.0486-0.012 = 0.0366 M

2.

                     NH4HS(s) <=====. NH3(g) + H2S(g)

at equi         0.296 M            0.0134 M   0.0134 M

after reestablished                        0.0232-X    0.0134 -X

KC =[NH3][H2S] = 0.0134^2 = 1.8*10^-4

1.8*10^-4 = (0.0232-X)(0.0134 -X)

X = 0.004 M

after reestablished

[NH3] = 0.0232-0.004 = 0.0192 M

[H2S] = 0.0134-0.004 = 0.0094 M

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