We will be looking at: A. Judging the extent of a reaction B. Predicting the direction of a reaction C. Calculating equilibrium concentrations

Transcription

1 13.5 Using the Equilibrium Constant We will be looking at: A. Judging the extent of a reaction B. Predicting the direction of a reaction C. Calculating equilibrium concentrations A. Judging the extent of a reaction: The magnitude of the constant (or K p ) gives an idea of the extent to which reactants are converted to products. Since the products are given in the numerator of the equilibrium constant equation, and the reactants are in the denominator, we can make general statements about the completeness of a given equilibrium reaction based on the value K. 1. If the constant is much larger than 10 3, this means the products dominate the reaction, and the reaction proceeds nearly to completion. H (g) + O (g) H O (g) HO 47.4x10 At 500K H O. If the constant is much smaller than 10-3, this means the reactants dominate the reaction, and the reaction proceeds nearly not at all. H O (g) H (g) + O (g) H O 48 ' 4.1x 10 At 500K H O 3. If the constant is between10-3 and 10 3, this means that neither the reactants or products dominate the reaction, and appreciable concentrations of all species are present in the equilibrium mixture. H (g) + I (g) HI (g) HI 57.0 At 700K H I 1

2 B. Predicting the direction of a reaction: the Reaction Quotient What happens if you substitute the concentrations of the products and reactants into the equilibrium constant equation when the system is not at equilibrium? You would get a value that does not equal the equilibrium constant. You know the system is not at equilibrium! But which way is the reaction going? Let s define the reaction quotient, Q c in exactly the same way we define the equilibrium constant. So for a general reaction aa + bb cc + dd with the constant Q c C A c a D B d b if Q c then the system is not yet at equilibrium, and the reaction is still proceeding to equilibrium. If Q c > this means we have more products and less reactants than the equilibrium mixture would have (since we are looking at products over reactants) and the reaction needs to create more reactants (and use up products) to get to equilibrium, so the reaction will be going from right to left. If Q c < this means we have more reactants and less products than the equilibrium mixture would have and the reaction needs to create more products (and use up reactants) to get to equilibrium), so the reaction will be going from left to right.

3 H (g) + I (g) HI (g) HI 57.0 At 700K H I If we had at a given moment in time, [H ] t = 0.80 mol/l, [I ] t = 0.5 mol/l, and [HI] t = 10.0 mol/l, then the reaction quotient, Q, would be Q c HI H I (10.0) (0.80)(0.5) 500 First, we see that the reaction quotient does not equal the equilibrium constant, so the system is not at equilibrium. Secondly, since Q > K the reaction will proceed from right to left, decomposing HI to form hydrogen and iodine until the system reaches equilibrium with K = 57. Problem 13.9 The equilibrium constant for the reaction NO (g) + O (g) NO (g) is 6.9 x 10 5 at 500 K. A 5.0 L reaction vessel at this temperature was filled with mol of NO, 1.0 mol of O, and 0.80 mol of NO. a) Is the reaction mixture at equilibrium? If not, which direction does the reaction proceed? b) What is the direction of the reaction if the initial amounts are 5.0 x 10-3 mol of NO, 0.0 mol of O, and 4.0 mol of NO? ANS: Solution done in class 3

4 C. Calculating equilibrium concentrations Once it is known that 1) a system is not at equilibrium, ) the direction in which the reaction must proceed in order to reach equilibrium is determined from a comparison of Q with K eq then the final equilibrium concentrations or partial pressures may be found. Problem: 0.30 mol each of Cl (g) and PCl 3 (g) and 3.0 mol of PCl 5 (g) are placed in a 1.0 L flask. At the temperature of the experiment, = 4.00 x 10 - for the reaction PCl 5 (g) < > PCl 3 (g) + Cl (g). Is the system at equilibrium? If not, in which direction must it proceed to reach equilibrium and what are the equilibrium concentrations? PCl 5 (g) < > PCl 3 (g) + Cl (g) [PCl 3 ] = [Cl ] = 0.30 M and [PCl 5 ] = 3.0M Q c = [PCl 3 ] [Cl ] / [PCl 5 ] =(0.30) (0.30) /3.0 = 0.030= 3.00 x 10 - Problem (continued) Q < K so reaction is not at equilibrium and must proceed > to attain equilibrium. Now we know system is not at equilibrium we can find equilibrium concentrations as before by setting up the appropriate table: [PCl 5 ] [PCl 3 ] [Cl ] initial [ ] change -x +x +x final [ ] equil 3.0-x 0.30+x 0.30+x Substitute into the equation for = [PCl 3 ] [Cl ] / [PCl 5 ] 4.0 x10 - = (0.30 +x) / (3.0 -x) rearranging gives x x x10 - = 0 This is a quadratic equation of the form ax + bx + c = 0, which has solutions given by b b 4 ac x a 4

5 Problem (continued) solving for x = (we use the positive x because we know that the reaction must go to the ) now calculate [ ] equil as set up in the table above [PCl 5 ] = x =.95 4 M [PCl 3 ] = [Cl ] = x = M Problem The equilibrium constant K p for the reaction C(s) + H O (g) CO (g) + H (g) is.44 at 1000 K. What are the equilibrium partial pressures of H O, CO, and H if the initial partial pressures are P HO = 1.0 atm, P CO = 1.00 atm, and P H = 1.40 atm? Ans: solution done in class 5

6 This document was created with WinPDF available at The unregistered version of WinPDF is for evaluation or non-commercial use only.