Unit XII: Disturbing Chemical Equilibrium

1. Unit XII: Disturbing Chemical Equilibrium … Chapter 16…

2. Relationship Between Kc and Kp • Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes where Kp = Kc (RT)n n = (moles of gaseous product) - (moles of gaseous reactant)

3. Relationship Between Kc and Kp 2Cl2(g) + 2H2O(g) 4HCl(g) + O2(g) Kp = 0.0752 at 480°C Kp = PHCl4·PO2 PCl22·PH2O2 Kc = ? Kp = Kc(RT)n 0.0752 = Kc(0.0821∙753)1 0.0752 = Kc(61.8213) Kc = 1.22 x 10 -3

4. Disturbing a Chemical Equilibrium

5. Disturbing a Chemical Equilibrium • Equilibrium may be disturbed 3 ways: • By changing the concentration of a reactant or product • By changing the volume (for systems of gases) • By changing the temperature

6. Disturbing a Chemical Equilibrium • By changing the concentration of a reactant or product • If the concentration of a reactant or product is changed from its equilibrium value at a given temperature, equilibrium will be reestablished eventually • New equilibrium concentrations will be different, but the equilibrium constant will be the same

7. Disturbing a Chemical Equilibrium • Assume equilibrium has been established in a 1.00-L flask with [butane] = 0.500 mol/L and [isobutane] = 1.25 mol/L Butane Isobutane K = 2.5 Then 1.50mol of butane is added. What are the concentrations of butane and isobutane when equilibrium is reestablished?

8. Disturbing a Chemical Equilibrium Butane Isobutane I: 0.500 1.25 (I’) 0.500+1.50 1.25 C: -x +x E: 2.00 – x 1.25 + x

9. Disturbing a Chemical Equilibrium K = [isobutane]/[butane] = 2.5 (1.25 +x)/(2.00-x) = 2.5 2.5(2.00 –x) = 1.25 + x 5 – 2.5x = 1.25 + x 3.75 = 3.5 x X = 1.07 mol/L [butane] = (0.500 +1.50 – x) = 0.93M [isobutane] = 1.25 + 1.07 = 2.32M

10. Disturbing a Chemical Equilibrium • By changing the volume (for systems of gases • For rxns involving gases, the stress of a volume decrease (increase in pressure) will be counterbalanced by a change in the equilibrium composition to one having a smaller number of gas molecules • For a volume increase (decrease in pressure), the equilibrium will favor the side of the rxn with the larger number of gas molecules • For a rxn in which there is no change in the number of gas molecules, a volume change will have no effect

11. Disturbing a Chemical Equilibrium • The formation of ammonia from its elements is an important industrial process: 3H2(g) + N2(g) 2NH3(g) • How does the equilibrium composition change when extra H2 is added? • Shifts it to the right • When extra NH3 is added? • Shifts it to the left • What is the effect on the equilibrium when the volume of the system is increased? • Shifts it the left

12. Disturbing a Chemical Equilibrium • By changing the temperature • Difficult because equilibrium changes with temperature • If you know whether an experiment is endothermic or exothermic you can make a qualitative prediction • When the temperature of a system at equilibrium increases, the equilibrium will shift in the direction that absorbs energy as heat • If the temperature decreases, the equilibrium will shift in the direction that releases energy as heat – that is, in the exothermic direction. • Changing the temperature changes the value of K

13. Disturbing a Chemical Equilibrium 2NOCl(g) 2NO(g) +Cl2(g) ΔrH°= +77.1kJ Does the equilibrium concentration of NOCl increase or decrease as the temperature of the system is increased? • It decreases

14. Disturbing a Chemical Equilibrium • Catalysts • Catalysts increase the rate of both the forward and reverse reactions • When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered.

15. Disturbing a Chemical Equilibrium 6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ΔH°= +2816kJ • Increase the [O2] • Increase the [CO2] • Decrease the [H2O] • Remove some of the C6H12O6 • Increase the pressure: in the system • By compression • Adding an inert gas • Add a catalyst Shifts left Shifts right Shifts left No Shift Shifts right No Shift No Shift