Chapter 14 Chemical Equilibrium K Chemistry part 2

Chapter 14: Chemical Equilibrium “K” Chemistry (part 2 of 3)

Reversible Reactions • Reversible reaction – can produce both the forward and reverse reactions… H 2(g) + I 2(g) 2 HI(g) • What happens initially? ? What direction is favored at the very beginning?

Dynamic Equilibrium • Eventually the rate of the forward reaction will balacne out and equal the rate of the reverse reaction… this reversible reaction is said to be in Dynamic Equilibrium (goes back and forth, …. ) • Dynamic Equilibrium – “for a chemical reaction the condition in which the rate of the forward reaction equals the rate of the reverse reaction”

The Equilibrium Constant (K) • The rates for the forward and reverse reactions are the same… NOT the concentrations!! • The Equilibrium Constant (K) – IT’S A CAPITAL K!! – defined as a ratio (at equilibrium) of the [products] raised to their stoichiometric coefficients divided by the [reactants] raised to their own coefficients – “Products over Reactants”

The Equilibrium Constant (K) •

Example •

Practice • Express the equilibrium constant for the following chemical equation CH 3 OH(g) CO(g) + 2 H 2(g)

Magnitude of K •

Magnitude of K • K<<1 Reverse direction is favored; forward reaction does not proceed very far • K ≈ 1 Neither direction is favored; forward reaction proceeds about halfway • K>>1 Forward reaction is favored; forward reaction proceeds essentially to completion

Conceptual Connection 14. 1 The equilibrium constant for the reaction A(g) B(g) Is 10. A reaction mixture initially contains 11 mol of A and 0 mol of B in a fixed volume of 1. 0 L. When equilibrium is reached, which of the following is true? A. ) The reaction mixture will contain 10 mol of A and 1 mol of B B. ) The reaction mixture will contain 1 mol of A and 10 mol of B C. ) The reaction mixture will contain equal amounts of A and B Answer: B. [B]/[A] = 10

Keq and the Chemical Equation • If you reverse the reaction, invert the equilibrium constant (take the reciprocal) • If you multiply the coefficients in the equation by a factor, raise the K by that same factor • If you add two or more equations together, multiply the corresponding equilibrium constants by each other to obtain the overall equilibrium constant

Practice • Consider the following chemical equation and equilibrium constant at 25 ⁰ C: 2 COF 2(g) CO 2(g) + CF 4(g) K = 2. 2 x 106 • Compute the equilibrium constant for the following reaction at 25⁰C CO 2(g) + CF 4(g) 2 COF 2(g) K’ = ? ?

Expressing Equilibrium Using Pressures • So far, we have only written K’s using [conc] of reactants and products • The partial pressures of gases are also proportional to its concentration… so we can write out an equilibrium expression in terms of partial pressures

Kc Vs. Kp •

Practice • Write out the equilibrium expression constant (Kc) for the following: Ca. CO 3(S) Ca. O(s) + CO 2(g) • What would be the expression using partial pressures?

More Practice • Write an equilibrium expression (Kc) for the following chemical equation: 4 HCl(g) + O 2(g) H 2 O(l) + 2 Cl 2(g) • What would be the expression using partial pressures?

Additional Info (for Test!!) • Read through and be able to do: – Comparing Kc and Kp mathematically • Kp = Kc(RT)Δn – “ICE Tables” • Page 630 – 633 – Reaction Quotients • How do you calculate them? • What do they mean? / Tell us?