1. 15.1c: Using ICE Tables Predicting Equilibrium Concentrations

2. Example 1 At a certain temperature a 1.00 L flask initially contained 0.298 mol PCl 3 (g) and 8.70x10 -3 mol PCl5(g). After the system had reached equilibrium, 2.00x10 -3 mol Cl 2 (g) was found in the flask. PCl 5 (g)  PCl 3 (g) + Cl 2 (g) Calculate the equilibrium concentrations of all the species and the value of K.

3. Example 1 PCl 5 (g)  PCl 3 (g) + Cl 2 (g) PCl 5 (g)  PCl 3 (g) + Cl 2 (g) I 8.70x10 -3 0.2980 C-x+x+x E 8.70x10 -3 -x = (8.70-2.00) x10 -3 = 6.70x10 -3 0.298+x = 0.298+2.00x10 -3 = 0.300 x = 2.00x10 -3

4. Approximations if K is very small, we can assume that the change (x) is going to be negligible can be used to cancel out when adding or subtracting from a “normal” sized number used to simplify algebra 0

5. Example 2 At 35°C, K=1.6x10 -5 for the reaction 2NOCl (g) ⇄ 2NO (g) + Cl 2 (g) Calculate the concentration of all species at equilibrium for the following mixtures a) 2.0 mol NOCl in 2.0 L flask b) 1.0 mol NOCl and 1.0 mol NO in 1.0 L flask c) 2.0 mol NOCl and 1.0 mol Cl 2 in 1.0 L flask

6. Example 2a 2.0 mol NOCl in 2.0 L flask [NOCl]=1.0 - (2 x 0.016)=0.97 M = 1.0 M, [NO]=0.032 M, [Cl 2 ]=0.016 M 2NOCl (g) ⇄ 2NO (g) + Cl 2 (g) I1.000 C-2x+2x+x E1.0-2x2xx

7. Example 2b 1.0 mol NOCl and 1.0 mol NO in 1.0 L flask [NOCl]=1.0 - (2x1.6x10 -5 )=0.999968 M = 1.0 M, [NO]=1.0 +(2x1.6x10 -5 )= 1.0 M, [Cl 2 ]=1.6x10 -5 M 2NOCl (g) ⇄ 2NO (g) + Cl 2 (g) I1.01.00 C-2x+2x+x E1.0-2x1.0+2xx

8. Example 2c 2.0 mol NOCl and 1.0 mol Cl 2 in 1.0 L flask [NOCl]= 2.0 - (2x4.0x10 -3 )=1.992 M = 2.0 M, [Cl 2 ]=1.0+4.0x10 -3 =1.0 M, [NO]=0.0080 M, 2NOCl (g) ⇄ 2NO (g) + Cl 2 (g) I2.001.0 C-2x+2x+x E2.0-2x2x1.0+x

9. Textbook: p 688 #6,9,10 LSM 15.1D