1 “ICE” CALCULATIONS

2 Typical Calculations PROBLEM: Place 1.00 mol each of H 2 and I 2 in a 1.0 L flask. Calc. equilibrium concentrations. H 2 (g) + I 2 (g)  2 HI(g)

3 H 2 (g) + I 2 (g)  2 HI(g) K c = 55.3 Step 1. Set up ICE table [H 2 ][I 2 ][HI] Initial Change Equilib

4 [H 2 ][I 2 ][HI] Initial Change-x-x+2x Equilib1.00-x1.00-x2x where x is defined as am’t of H 2 and I 2 consumed on approaching equilibrium. Step 2. define “X”

5 H 2 (g) + I 2 (g) e 2 HI(g) K c = 55.3 Step 3. Put equilibrium concentrations into K c expression.

6 [H 2 ] = [I 2 ] = x = 0.21 M [HI] = 2x = 1.58 M [H 2 ] = [I 2 ] = x = 0.21 M [HI] = 2x = 1.58 M Step 4. Solve K c expression - take square root of both sides. x = 0.79 Therefore, at equilibrium H 2 (g) + I 2 (g) ---> 2 HI(g) K c = 55.3

7 Nitrogen Dioxide Equilibrium N 2 O 4 (g) ---> 2 NO 2 (g) 

8 If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an ICE table [N 2 O 4 ][NO 2 ] Initial0.500 Change Equilib

9 Step 2. Define “X” [N 2 O 4 ] [NO 2 ] Initial Change-x+2x Equilib x 2x Nitrogen Dioxide Equilibrium N 2 O 4 (g) ---> 2 NO 2 (g)

10 Step 3. Substitute into K c expression and solve. Rearrange: ( x) = 4x x = 4x 2 4x x = 0 This is a QUADRATIC EQUATION ax 2 + bx + c = 0 a = 4b = c =

11 Solve the quadratic equation for x. ax 2 + bx + c = 0 a = 4b = c = x = ± 1/8(0.046) 1/2 = ± = ± 0.027

12 Nitrogen Dioxide Equilibrium N 2 O 4 (g) ---> 2 NO 2 (g) x = or negative value is not reasonable. Conclusion: x = M [N 2 O 4 ] = x = 0.47 M [NO 2 ] = 2x = M OR x = ± 1/8(0.046) 1/2 = ± 0.027

13 Step 2. Define “X” [N 2 O 4 ] [NO 2 ] Initial Change-x+2x Equilib x APPROXIMATE: K is small: x will be small

14 Step 3. Substitute into K c expression and solve. Rearrange: (0.50) = 4x = 4x /4 = x 2 X = Step 4. check if assumption is correct 0.50 – = 0.47 same?

15 Solving Quadratic Equations Recommend you solve the equation exactly on a calculator