Chemical Equilibrium 14.4-14.7 Charlie Derbyshire

Equilibrium Constant in Gaseous Reactions The equilibrium constant a proportion of the concentration of the reactants and products. However, gaseous reactions have concentrations proportional to the partial pressures of each particular gas, so the equilibrium constant can also express equilibrium in terms of partial pressure. So, with equilibrium constant for concentration being Kc and the equilibrium constant for pressure being Kp, Kc=[C]c[D]d/[A]a[B]b and similarly, Kp=(Pc)c(Pd)d/(PA)a(PB)b, where P is the partial pressure of those individual gases in atmospheres. Because the equilibrium constant equals concentration/concentration or pressure/pressure, it has no units, only a numerical value.

Tying concentration with pressure Combining the law of mass action with the ideal gas law, we can determine Kp in terms of Kc: Kp=Kc(RT)∆n ∆n is the sum of all of the coefficients of the products minus the sum of the coefficients of the reactants Kp=Kc when the moles of the reactants=moles of the products because ∆n=0.

Exceptions to Finding Equilibrium Constant If a reaction forms a solid, that is excluded when finding the equilibrium constant because the concentration is constant no matter how much solid there is. If a reaction forms a pure liquid, that is also excluded when finding equilibrium constant for the same reason: adding more of that liquid does not change its concentration.

Practice A mixture of the reaction below at a constant temperature of 293K has [C2O]=0.105M, [H2]=0.114M, and [CH3OH]=0.185M. Find Kc and Kp. C2O(g) + 2H2(g) ⇌ CH3OH(g)+C(s) Kc=[CH3OH]/[H2]2[C2O] Kc=(0.185M)/(0.114M)2(0.105M) Kc=135.573 Kp=Kc(RT)∆n Kp=135.573((0.08206L atm mol-1K-1)(293K))1- (1+2) Kp=0.235

Using Limited Concentrations When finding the equilibrium constant, we only need to know the initial concentrations and one equilibrium concentration of the reactants. We can find the other equilibrium concentrations of the reactants with the stoichiometric coefficients and an ICE table (initial, change, equilibrium) A(g) ⇌ 2B(g) [A]0=0.25M [A]e=0.1M [B]0=0.5M Because A yields 2B, ∆ [B] = -2∆[A] ICE table [A] [B] Initial 0.25M 0.5M Change -0.15M +0.3M Equilibrium 0.1M 0.8M

The Reaction Quotient In order to determine which direction of change that the reaction will go to reach equilibrium, we use the reaction quotient, or Qc(Qp when using pressure). Qc=[C]c[D]d/[A]a[B]b Qp=PcCPdD/PaAPbB These are the same equations as for the equilibrium constant, but are used in any point of the reaction, not just at equilibrium. When there are only reactants in a mixture, Qc=0, and when there are only products, Qc=∞.

Using Reaction Quotient to Find Direction of Change We find which direction of change the reaction is going by comparing Q to K. If Q<K, the reaction goes from the reactants towards the products If Q>K, the reaction goes from the products towards the reactants(reverse reaction) If Q=K, the reaction is at equilibrium If Q=4.2 and K=3, would the reaction go in forward or reverse direction?