I. The Reaction Quotient We can write K only for a reaction at equilibrium

1. I. The Reaction Quotient • We can write K only for a reaction at equilibrium • How can we describe a reaction at other times? N2 + 3H2 2NH3 • Q = reaction quotient has the same form as K, but is at non-equilibrium times. • Q can tell us how a reaction will change to get to equilibrium • If Q = K, we are at equilibrium • If Q > K, then products > reactants and the reaction will shift left • If Q < K, then reactants > products and the reaction will shift right • Example: For the above reaction, K = 6 x 10-2. Predict change: • [N2]0 = 1 x 10-5 M, [H2]0 = 0.002 M, [NH3]0 = 0.001M • [N2]0 = 1.5 x 10-5 M, [H2]0 = 0.354 M, [NH3]0 = 0.0002M • [N2]0 = 5 M, [H2]0 = 0.01 M, [NH3]0 = 0.0001 M Describes the reaction when not at equilibrium

2. Solving Equilibrium Problems • Problems where equilibrium concentrations (or pressures) are given • Example: N2O4(g) 2NO2(g) • KP = 0.133 • P(N2O4) at equilibrium = 2.71 atm • Find P(NO2) at equilibrium • Example: PCl5(g) PCl3(g) + Cl2(g) • [PCl3]0 = 0.298 M, [PCl5]0 = 0.0087 M • At equilibrium, [Cl2]0 = 0.002 M • Calculate all concentrations and K • Procedure when not given equilibrium concentrations • Write the balanced equation • Write the equilibrium expression • List the initial conditions • Calculate Q to decide which way the reaction will shift • Write equilibrium conditions using x as unknown changes in concentration • Substitute the equilibrium conditions into the equilibrium expression and solve for x • Check your answer to see if it gives the correct value of K

3. Example: 3 mol of hydrogen and 6 mol of fluorine are mixed in a 3 L flask to give hydrogen fluoride gas. K = 115. Find the equilibrium concentrations. • Write the balanced equation: H2 + F2 2HF • Write the equilibrium expression: • List the initial conditions: [H2]0 = 1 M, [F2]0 = 2 M, [HF]0 = 0 • Calculate Q to decide which way the reaction will shift • Write equilibrium conditions using x as unknown changes in concentration

4. Shorthand: H2 + F2 2HF Initial: 1 2 0 Change -x -x +2x Equilibrium 1-x 2-x 2x • Substitute and solve for x • Check your answer to see if it gives the correct value of K a = 111, b = -345, c = 230 x = 2.14 M or x = 0.968 M Reject x = 2.14 M since [H2] = 1 – x = 1 – 2.14 = -1.14 M x = 0.968M [HF] = 2x = 1.936M [H2] = 1- x = 0.032M [F2] = 2- x = 1.032M

5. Example: CO(g) + H2O(g) CO2 + H2 K = 5.1 Find equilibrium conditions if 1 mol of each is mixed in a 1L flask. • Example: 3 mol of hydrogen and 3 mol of fluorine are mixed in a 1.5 L flask with 3 mol hydrogen fluoride gas. K = 115. Find the equilibrium concentrations. H2 + F2 2HF • Example: H2 + I2 2HI KP = 100. If we mix HI (P = 0.5 atm), H2 (P = 0.01 atm), and I2 (P = 0.005) in a 5 L flask, what are eq. conc.? • Approximations when x is small 2NOCl(g) 2NO + Cl2(g) K = 1.6 x 10-5 Initial 0.5 M 0 0 Equil. 0.5 – 2x 2x x Since K is small, x will be very small.

6. Le Chatelier’s Principle • We can change the position of an equilibrium by changing conditions • N2 + 3H2 2NH3 • Low temperature and high pressure favor the forward reaction • Le Chatelier’s Principle: When a chemical system at equilibrium is disturbed, the system shifts in a direction that minimizes the disturbance. • Effect of change in concentrations N2 + 3H2 2NH3 K = 0.0596 • At equilibrium, [N2] = 0.399 M, [H2] = 1.197 M, [NH3] = 0.202 M • Let’s add 1 M N2. How does the equilibrium shift? • Too much N2 for equilibrium • Reaction must adjust to get back to equilibrium • Reaction must shift to the right Eq. shifts to the right

7. We could do the x calculation and find the new concentrations if we wanted, but sometimes you just want to know which way the reaction will shift. Le Chatelier’s principle lets us do that simply. • Another Example: What would happen if we add N2O4?

9. Example: As4O6(s) + 6C(s) As4(g) + 6CO(g) • Add CO • Add or remove As4O6(s) • Remove As4(g) • Effect of Change in Pressure/Volume • If we add/remove reactant/product, we change concentration (see above) • Adding an inert gas, doesn’t change concentrations, so no effect on equilibrium • Changing the size of the flask: V↓ = P↑ V↑ = P↓

10. The volume changes, so the concentrations must change • To re-establish equilibrium, the concentrations will adjust • Fewer gas particles are favored at higher pressures • This reduces the pressure (Le Chat’s principle) • N2(g) + 3H2(g) 2NH3(g) Increase the pressure • There are 4 particles on the left and 2 on the right • To reduce the pressure, the equilibrium will shift right • If we increased the volume and decreased the pressure, the reaction will shift to increase the number of particles (shift left) • Example: Reduce the volume on the following reactions: • P4(s) + 6Cl2(g) 4PCl3(l) • PCl3(g) + Cl2(g) PCl5(g) • PCl3(g) + 3NH3(g) P(NH2)3(g) + 3HCl(g)

11. Effect of change in Temperature • The above changes (P, Concentration) effect the equilibrium position, but not the equilibrium constant K • Changing Temperature changes the equilibrium constant K • We can predict the changes by treating heat as a product or reactant of every reaction • Exothermic reactions produce heat as a product • Adding heat (increasing temp.) shifts away from heat, to left, K decreases • Removing heat (decreasing temp.) shifts towards heat, to right, K increases • N2 + 3H2 2NH3 + 93 kJ/mol (DH = - 93 kJ/mol) • Temperature (oC): 500 600 700 800 K: 90 3 0.3 0.04 • Endothermic reactions require heat as a reactant • Add heat = shift to right, K increases • Remove heat = shift to left, K decreases • 556 kJ/mol + CaCO3(s) CaO(s) + CO2(g) (DH = + 556 kJ/mol) 6. Examples: N2(g) + O2(g) 2NO(g) DH = 181 kJ/mol 2SO2(g) + O2(g) 2SO3DH = - 198 kJ/mol)