Equilibrium Constant and Shifting Equilibrium Notes

Equilibrium Constant

• The equilibrium-constant expression can be used to calculate the concentrations of reactants and products at equilibrium knowing the equilibrium constant.

• Example:

  • At 425°C, an equilibrium system contains 0.015 mol/L of H2 and I2.

  • To find [HI]: K = {[HI]^2 \over [H2][I2]}

  • Rearranging: [HI] = \sqrt{K[H2][I2]}

  • [HI] = \sqrt{54.3(0.015)(0.015)} = 0.11 mol/L

• Values for K change with temperature.

Sample Problem A

• Determining the equilibrium constant (K) for the system N2(g) + O2(g) \rightleftharpoons 2NO(g) at 1500 K.

  • Given: [N2] = 6.4 x 10^{-3} mol/L, [O2] = 1.7 x 10^{-3} mol/L, [NO] = 1.1 x 10^{-5} mol/L

  • Unknown: K

  • K = {[NO]^2 \over [N2][O2]}

  • K = {(1.1 x 10^{-5})^2 \over (6.4 x 10^{-3})(1.7 x 10^{-3})} = 1.1 x 10^{-5}

• A small K value is consistent with more N2 and O2 than NO at equilibrium.

Practice Problems

1. Calculating the equilibrium constant for N2(g) + 3H2(g) \rightleftharpoons 2NH3(g) at 500°C with given equilibrium concentrations: [N2] = 0.602 mol/L, [H2] = 0.420 mol/L, [NH3] = 0.113 mol/L

2. Calculating K for AB2C(g) \rightleftharpoons B2(g) + AC(g) at 900 K in a 5.00-L vessel with equilibrium amounts 0.084 mol AB2C, 0.035 mol B2, and 0.059 mol AC.

3. Calculating K for SO2(g) + O2(g) \rightleftharpoons SO3(g) at 600°C with equilibrium concentrations: [SO2] = 1.50 mol/L, [O2] = 1.25 mol/L, [SO3] = 3.50 mol/L

Section 1 Formative Assessment

1. Chemical equilibrium is the dynamic state where the rate of the forward reaction equals the rate of the reverse reaction.

2. An equilibrium constant (K) is the ratio of product to reactant concentrations at equilibrium.

3. The equilibrium constant value indicates the relative amounts of products and reactants at equilibrium. A very large K indicates that the products are favored, while a very small K indicates that the reactants are favored.

4. A chemical equilibrium expression is the mathematical expression relating reactant and product concentrations at equilibrium to the equilibrium constant.

5. For the reaction HCl(aq) + H2O(l) \rightleftharpoons H3O^+(aq) + Cl^-(aq), the value of K should be very large because HCl is a strong acid that dissociates completely in water

6. 4HCl(g) + O2(g) \rightleftharpoons 2Cl2(g) + 2H2O(g); K = {[Cl2]^2[H2O]^2 \over [HCl]^4[O2]}

7. H2(g) + Cl2(g) \rightleftharpoons 2HCl(g); K = {[HCl]^2 \over [H2][Cl2]} = {(0.0625)^2 \over (0.00450)(0.00450)} = 192.9

8. H2(g) + I2(g) \rightleftharpoons 2HI(g); K = {[HI]^2 \over [H2][I2]} = {(1.77 x 10^{-2})^2 \over (1.83 x 10^{-3})(3.13 x 10^{-3})} = 54.3

9. For H2(g) + I2(g) \rightleftharpoons 2HI(g) at 425°C, with [H2] = [I2] = 4.79 x 10^{-4} mol/L and K = 54.3; [HI] = \sqrt{K[H2][I2]} = \sqrt{54.3(4.79 x 10^{-4})(4.79 x 10^{-4})} = 1.11 x 10^{-3} mol/L

10. For 2HI(g) \rightleftharpoons H2(g) + I2(g), at 425°C, K = {[H2][I2] \over [HI]^2}, The value obtained will be the inverse of the value obtained earlier since the reaction is reversed

Fixing the Nitrogen Problem

• The chemical industry synthesizes nitrogenous fertilizers, which increased agricultural production.

• Prior to 1915, people relied on natural resources for fertilizer, leading to fears of world starvation due to dwindling supplies.

• The

The chemical industry synthesizes nitrogenous fertilizers, which increased agricultural production.

Prior to 1915, people relied on natural resources for fertilizer, leading to fears of world starvation due to dwindling supplies.

The Haber-Bosch process was developed to produce ammonia from nitrogen and hydrogen, enabling a large-scale production of fertilizers.

This process uses high temperatures and pressures to combine nitrogen from the atmosphere with hydrogen derived primarily from natural gas.

The availability of nitrogen fertilizers has significantly boosted crop yields, helping to support the growing global population. However, excessive use of fertilizers can lead to environmental issues such as water pollution and soil degradation.