Question

The equilibrium constant, Kc, for the following reaction is 1.05×10-3 at 446 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently large sample of PCl5(g) is introduced into an evacuated vessel at 446 K, the equilibrium concentration of Cl2(g) is found to be 0.498 M. Calculate the concentration of PCl5 in the equilibrium mixture. ____M _______________________________________________________________________________________________________________ A student ran the following reaction in the laboratory at 689 K: N2(g) + 3H2(g) 2NH3(g) When she introduced 4.06×10-2 moles of N2(g) and 5.20×10-2 moles of H2(g) into a 1.00 liter container, she found the equilibrium concentration of H2(g) to be 5.00×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc = _______________________________________________________________________________________________________________________________ A student ran the following reaction in the laboratory at 658 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 8.28×10-2 moles of NH3(g) into a 1.00 liter container, she found the equilibrium concentration of NH3(g) to be 6.82×10-3 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =

Answer 1

Question 1.

The equilibrium constant, Kc, for the following reaction is 1.05×10-3 at 446 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently large sample of PCl5(g) is introduced into an evacuated vessel at 446 K, the equilibrium concentration of Cl2(g) is found to be 0.498 M. Calculate the concentration of PCl5 in the equilibrium mixture. ____M

The Kc expression --> Kc = [PCl3][Cl2]/[PCl5]

in equilibrium

[Cl2] = [PCl3] due t sotichiometry, i.e. 1:1

[Cl2] = [PCl3] = 0.498

Then, PCl5 = x

Kc = [PCl3][Cl2]/[PCl5]

1.05*10^-3 = 0.498*0.498/(x)

x = (0.498*0.498)/(1.05*10^-3)

x = 236.19 M for PCl5

[PCl5] = 236.19 M

Question 2.

A student ran the following reaction in the laboratory at 689 K: N2(g) + 3H2(g) 2NH3(g) When she introduced 4.06×10-2 moles of N2(g) and 5.20×10-2 moles of H2(g) into a 1.00 liter container, she found the equilibrium concentration of H2(g) to be 5.00×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =

The Kc expression, Kc = [NH3]^2 / ([N2][H2]^3)

initially

[N2] = mol/V = 4.06*10^-2 / 1 = 4.06*10^-2

[H2] = mlo/V = 5.2*10^-2 / 1 = 5.2*10^-2

[NH3] = 0

in equilbirium

[N2] = 0.0406 - x

[H2] = 0.0520 -3x

[NH3] = 0 +2x

we know that [H2] in eq = 0.05

[H2] = 0.0520 -3x = 0.05

x = (0.050 - 0.052)/(-3) = 0.0006667

[N2] = 0.0406 - 0.0006667 = 0.0399333

[H2] = 0.0520 -3*0.0006667 = 0.0499999

[NH3] = 0 +2*0.0006667 = 0.0013334

Now, substitute in Kc

Kc = [NH3]^2 / ([N2][H2]^3)

Kc = (0.0013334^2)/((0.0399333 )(0.0499999^3))

Kc = 0.3561

Question 3.

A student ran the following reaction in the laboratory at 658 K: 2NH3(g) N2(g) + 3H2(g) When she introduced 8.28×10-2 moles of NH3(g) into a 1.00 liter container, she found the equilibrium concentration of NH3(g) to be 6.82×10-3 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc

The Kc expression

Kc = [N2][H2]^3 /[NH3]^2

[N2] = 0

[H2] 0

[NH3] = 8.28*10^-2 = 0.0828 M

in equilibrium

[N2] = 0 + x

[H2] 0 +3x

[NH3] = 0.0828 -2x

and we know

[NH3] = 0.0828 -2x = 6.82*10^-3

x =  (6.82*10^-3 -0.0828 )/(-2) = 0.03799

[N2] = 0 + x = 0.03799

[H2] 0 +3x = 3*0.03799 = 0.11397

[NH3] = 0.0828 -2x =0.0828 -2*0.03799 = 0.00682 M

Kc = (0.03799)(0.11397^3)/(0.00682 ^2)

Kc = 1.209127