Question

1.) The equilibrium constant for the chemical equation

N_2(g)+3H_{2(g) <-->} 2NH_3(g)

is Kp = 1.09 at 209 °C. Calculate the value of the Kc for the reaction at 209 °C.

2.) At a certain temperature, 0.3411 mol of N2 and 1.581 mol of H2 are placed in a 1.50-L container.

N_2(g)+3H_{2(g) <-->} 2NH_3(g)

At equilibrium, 0.1801 mol of N2 is present. Calculate the equilibrium constant, Kc.

3.) At a certain temperature, the Kp for the decomposition of H2S is 0.748.

H₂S_(g) <--> H_2(g)+S_(g)

Initially, only H2S is present at a pressure of 0.189 atm in a closed container. What is the total pressure in the container at equilibrium?

4.) An equilibrium mixture contains 0.600 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container.

CO_(g)+H₂O(g) <--> CO_2(g) + H_2(g)

How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol once equilibrium has been reestablished?

Answer 1

1)

kp = kc x R x T^(dn)

dn is the difference in mol = 2-(3+1) = -2

T= 209 + 273 = 482 °K

kc= 1.09 / (0.0821 x 482)^-2

kc = 1706.9

2)

[N2] = 0.3411 mol /1.5 L = 0.2274 M

[H2] = 1.581 mol /1.5 L = 1.054 M

---

[N2]eq = 0.1801 mol / 1.5 L = 0.1200 M

N2(g) + 3H2 <----> 2NH3 (g)

0.2274 | 1.054 | 0

-x | -3x | +2x

0.2274-x| 1.054-3x| 2x

[N2]eq = 0.2274 - xM = 0.1200 --> x = 0.1073 M

[H2]eq = 1.054 - (3 x 0.1073) = 0.7320 M

[NH3]eq = 2 x 0.1073 M = 0.2146 M

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

Kc = 4.60e-2 / 0.1200 x 0.3922

Kc = 0.9784

3)

H2S <---> H2 + S

0.189 | 0 | 0

-x | +x | +x

0.189 - x | x | x

Kp = [H2][S]/[H2S] = x^2/(0.189 - x)

x2 + 0.748x - 0.0815 = 0

x = 0.0965

Total pressure = 0.189 - x + x + x = 0.189 + 0.0956 = 0.2855 atm