Chemical Equilibrium PART 2

Calculating Equilibrium Concentrations 1. When all but one equilibrium concentration and the value of Kc are known. 2. When the value of Kc and the initial concentrations are known. When the Kc expression is a perfect square: solving a linear equation. When the Kc expression is not a perfect square: solving a quadratic equation.

Example to consider; Nickel(II) oxide can be reduced to the metal by treatment with carbon monoxide. CO(g) + NiO(s) ⇌ CO2(g) + Ni(s) If the partial pressure of CO is 100. mmHg and the total pressure of CO and CO2 does not exceed 1.0 atm, will this reaction occur at 1500 K at equilibrium? (Kp = 700. at 1500 K.)

Nitrogen and oxygen form nitric oxide Nitrogen and oxygen form nitric oxide. N2(g) + O2(g) 2NO(g) If an equilibrium mixture at 25°C contains 0.040 M N2 and 0.010 M O2, what is the concentration of NO in this mixture? The equilibrium constant at 25°C is 1.0 × 10-30.

When the initial concentration and the value of Kc are known, we return to the stoichiometric chart of initial, change, and equilibrium (ICE) amounts or concentrations to find the equilibrium concentrations.

Problem; Hydrogen iodide decomposes to hydrogen gas and iodine gas. 2HI(g) H2(g) + I2(g) At 800 K, the equilibrium constant, Kc, for this reaction is 0.016. If 0.50 mol HI is placed in a 5.0-L flask, what will be the composition of the equilibrium mixture in molarities?

2HI(g) H2(g) + I2(g) Initial Change Equilibrium

The Kc expression is, Kc = Substituting:

When the Kc expression is not a perfect square, the equation must be rearranged to fit the quadratic format: ax2 + bx + c = 0 The solution is

Consider; N2O4 decomposes to NO2 Consider; N2O4 decomposes to NO2. The equilibrium reaction in the gas phase is N2O4(g) 2NO2(g) At 100°C, Kc = 0.36. If a 1.00-L flask initially contains 0.100 M N2O4, what will be the equilibrium concentration of NO2?

Again, we begin with the table: N2O4(g) 2NO2(g) Initial Change Equilibrium

Given: H2(g) + F2(g) 2HF(g); Kc = 1. 15 × 102 3 Given: H2(g) + F2(g) 2HF(g); Kc = 1.15 × 102 3.000 mol of each species is put in a 1.500-L vessel. What is the equilibrium concentration of each species?

The value of Kc at 227°C is 0.0952 for the following reaction: CH3OH(g) CO(g) + 2H2(g) What is Kp at this temperature?

Le Châtelier’s Principle When a system in chemical equilibrium is disturbed by a change in temperature, pressure, or concentration, the system shifts in equilibrium composition in a way that tends to counteract this change of variable.

When a substance that is part of the equilibrium is added to the mixture, the equilibrium shifts to use it (in a direction that makes the substance a reactant). When a substance that is part of the equilibrium is removed from the mixture, the equilibrium shifts to produce it (in a direction that makes the substance a product). Changes in the partial pressure of substances that are part of the equilibrium are handled in the same way as adding or removing a substance.

The following reaction is at equilibrium: COCl2(g) CO(g) + Cl2(g) a. Predict the direction of reaction when chlorine gas is added to the reaction mixture. b. Predict the direction of reaction when carbon monoxide gas is removed from the mixture.

COCl2(g) CO(g) + Cl2(g) When we add Cl2, the reaction will shift in the reverse direction to use it. Note: reverse = left = . When we remove CO, the reaction will shift in the forward direction to produce it. Note: forward = right = .

A change in the total pressure occurs because of a change in the volume of the reaction container. When the size of the container decreases, the overall pressure increases. The reaction will shift to reduce the pressure—that is, it will shift toward the side of the reaction with fewer gas molecules.

When the size of the container increases, the overall pressure decreases. The reaction will shift to increase the pressure—that is, it will shift toward the side with more gas molecules. In the event that both sides of the equilibrium reaction have the same number of moles of gas, pressure has no effect on the equilibrium.

In which direction will each reaction shift when the volume of the reaction container is increased? CO(g) + 2H2(g) CH3OH(g) 2SO2(g) + O2(g) 2SO3(g) c. COCl2(g) CO(g) + Cl2(g)

Changing the temperature changes the value of the equilibrium constant Changing the temperature changes the value of the equilibrium constant. Changing the temperature can also cause a shift in the equilibrium. The direction of each of these changes depends on the sign of DHo.

For an endothermic reaction, DHo > 0 (positive), we consider that heat is a reactant. For an exothermic reaction, DHo > 0 (negative), we consider that heat is a product.

For an endothermic reaction, increasing the temperature increases the value of Kc. For an exothermic reaction, increasing the temperature decreases the value of Kc. Decreasing the temperature has the opposite effect.

In addition to the value of Kc, we can consider the direction in which the equilibrium will shift. When heat is added (temperature increased), the reaction will shift to use heat. When heat is removed (temperature decreased), the reaction will shift to produce heat.

Given: 2H2O(g) 2H2(g) + O2(g); DHo = 484 kJ Would you expect this reaction to be favorable at high or low temperatures? We rewrite the reaction to include heat: Heat + 2H2O(g) 2H2(g) + O2(g) When heat is added, the reaction shifts forward = right = . The reaction is favorable at high temperatures.