2. CALCULATIONS GIVEN K eq Example #1 CH 4(g) + H 2 O (g) <===> CO (g) + 3 H 2(g) At 1500°C, K eq = 5.67 [CO] = 0.300 M, [H 2 ] = 0.800 M and [CH 4 ] = 0.400 M What is [H 2 O] at equilibrium?
3. CALCULATIONS GIVEN K eq Example #1 CH 4(g) + H 2 O (g) <===> CO (g) + 3 H 2(g) At 1500°C, K eq = 5.67, [CO] = 0.300 M, [H 2 ] = 0.800 M and, [CH 4 ] = 0.400 M What is [H 2 O] at equilibrium? K eq = [CO][H 2 ] 3 [CH 4 ][H 2 O] [H 2 O] = [CO][H 2 ] 3 [CH 4 ] K eq [H 2 O] = [0.300M][0.800M] 3 [0.400M] 5.67 [H 2 O] = 0.0677M .: [H 2 O] at equilibrium was 6.77x10 -2 M
4. CALCULATIONS GIVEN K eq Example #2 N 2(g) + 3 H 2(g) <===> 2 NH 3(g) What is [NH 3 ] when [N 2 ] = 0.45 M, [H 2 ] = 1.10 and Keq = 1.7 x 10 -2 ?
5. CALCULATIONS GIVEN K eq Example #2 N 2(g) + 3 H 2(g) <===> 2 NH 3(g) What is [NH 3 ] when [N 2 ] = 0.45 M, [H 2 ] = 1.10 M and Keq = 1.7 x 10 -2 ? K eq = [NH 3 ] 2 [N 2 ][H 2 ] 3 K eq [N 2 ][H 2 ] 3 = [NH 3 ] 2 0.017 [0.45M][1.10M] 3 = [NH 3 ] 2 0.01018215 = [NH 3 ] 2 √ 0.01018215 = √ [NH 3 ] 2 0.10090664 = [NH 3 ] .: the [NH 3 ] is 1.0x10 -1 M
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8. CALCULATIONS GIVEN K eq CO (g) + H 2 O (g) <=> CO 2(g) + H 2(g) K eq = [CO 2 ][H 2 ] [CO][H 2 O] 4.06 = _____ [x][x]______ [0.100-x][0.100-x] I 0.100 0.100 0 0 C –x -x +x +x E 0.100-x 0.100-x x x 4.06 [0.100-x][0.100-x] = x 2 (0.406-4.06x)[0.100-x] = x 2 0.0406 – 0.406x -0.406x + 4.06x 2 = x 2 3.06x 2 – 0.812x + 0.0406 = 0 Time to use the quadratic formula!
9. CALCULATIONS GIVEN K eq CO (g) + H 2 O (g) <=> CO 2(g) + H 2(g) I 0.100 0.100 0 0 C –x -x +x +x E 0.100-x 0.100-x x x 3.06x 2 – 0.812x + 0.0406 = 0 Quadratic equation: a x 2 + b x + c = 0 Quadratic formula: x = -b ± √(b 2 -4ac) 2a x = -(-0.812) ± √[-0.812 2 -4(3.06)(0.0406)] 2(3.06) x = 0.19853 or 0.066832 This is the correct answer. Otherwise you may end up with negative reactant concentrations
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12. CALCULATIONS GIVEN K eq Example #4 H 2(g) + I 2(g) <===> 2 HI (g) K eq = 55.6 If initial [H 2 ] = 0.200 M and initial [I 2 ] = 0.200 M. What is [HI] at equilibrium?
13. CALCULATIONS GIVEN K eq Example #4 If initial [H 2 ] = 0.200 M and initial [I 2 ] = 0.200 M. K eq = 55.6 What is [HI] at equilibrium? H 2(g) + I 2(g) <=> 2 HI (g) I 0.200 0.200 0 C -x -x +2x E 0.200-x 0.200-x 2x K eq = _ [HI] 2 [H 2 ][I 2 ] 55.6 = _ [HI] 2 [H 2 ][I 2 ] 55.6 = ______ [2x] 2 _______ [0.200-x][0.200-x]
14. CALCULATIONS GIVEN K eq Example #4 If initial [H 2 ] = 0.200 M and initial [I 2 ] = 0.200 M. K eq = 55.6 What is [HI] at equilibrium? H 2(g) + I 2(g) <=> 2 HI (g) 55.6 = ______ [2x] 2 _______ [0.200-x][0.200-x] 55.6 = ___ 4x 2 ___ [0.200-x] 2 7.456 = ___ 2x ___ 0.200-x 1.4913 – 7.4565x = 2x 1.4913 = 9.4565x 0.1577 = x
15. CALCULATIONS GIVEN K eq Example #4 If initial [H 2 ] = 0.200 M and initial [I 2 ] = 0.200 M. K eq = 55.6 What is [HI] at equilibrium? H 2(g) + I 2(g) <=> 2 HI (g) x = 0.1577 I 0.200 0.200 0 C -x -x +2x E 0.200-x 0.200-x 2x 2x = 2 ( 0.1577) 2x = 0.3154M .: the equilibrium [HI (aq) ] = 3.15 x 10 -1 M
16. CALCULATIONS GIVEN K eq Example #5 CO (g) + H 2 O (g) <===> CO 2(g) + H 2(g) At equilibrium, K eq = 10.0. A reaction vessel is found to contain 0.80 M CO, 0.050 M H 2 O, 0.50 M CO 2 and 0.40 M H 2 . Determine if the reaction is at equilibrium.
17. CALCULATIONS GIVEN K eq Q When testing if conditions are at equilibrium Q is the symbol used, rather than K eq . Q is the “test K eq ” variable. Q = [products] [reactants]
18. CALCULATIONS GIVEN K eq Example #5 CO (g) + H 2 O (g) <===> CO 2(g) + H 2(g) At equilibrium, K eq = 10.0. A reaction vessel is found to contain 0.80 M CO, 0.050 M H 2 O, 0.50 M CO 2 and 0.40 M H 2 . Determine if the reaction is at equilibrium. Q = [CO 2 ][H 2 ] [CO][H 2 O] Q = [0.50][0.40] [0.80][0.050] Q = 5.0 Since k eq ≠ Q, then the reaction is not at equilibrium .: the reaction is not at equilibrium
![CALCULATIONS GIVEN K eq Example #1 CH 4(g) + H 2 O (g) <===> CO (g) + 3 H 2(g) At 1500°C, K eq = 5.67 [CO] = 0.300 M, [H 2 ] = 0.800 M and [CH 4 ] = 0.400 M What is [H 2 O] at equilibrium?](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)