Most chemical reactions are reversible. Chemical equilibrium can only occur in a closed system.
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· Most chemical reactions are reversible. · Chemical equilibrium can only occur in a closed system. · Equilibrium is when the rates are equal AND the [ ]s remains constant. · K c can be used to determine the equilibrium position.
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Most chemical reactions are reversible. Chemical equilibrium can only occur in a closed system. Equilibrium is when the rates are equal AND the [ ]s remains constant. K c can be used to determine the equilibrium position. K c and Equilibrium Problems. (4 types). - PowerPoint PPT Presentation
Transcript of Most chemical reactions are reversible. Chemical equilibrium can only occur in a closed system.
· Most chemical reactions are reversible.
· Chemical equilibrium can only occur in a closed system.
· Equilibrium is when the rates are equal AND the [ ]s remains constant.
· Kc can be used to determine the equilibrium position.
Kc and Equilibrium Problems
(4 types)
• Make equilibrium constant (K) calculations with various missing values.
• Use ICE tables to successfully complete
Equilibrium calculations.
*Always make sure the equation is balanced, FIRST*
At 225°C, a 2.0 L container holds 0.040 moles of N2, 0.15 moles of H2 and 0.50 moles of NH3. If the system is at equilibrium, calculate KC.
1. Change all into concentrations - mol/L
2. Write the equilibrium law for the reaction
N2(g) + 3 H2(g) 2 NH3(g)
0.040 mol N2
2.0 L= 0.020 M N2
= 0.075 M H2
= 0.25 M NH3
3. Substitute the concentrations and calculate K.(Note: no units for K)
The equilibrium concentrations of N2 and NO are 1.40 mol/L and 5.20 mol/L respectively. Calculate the KC.
According to Eq stoichiometry – the same amount of O2 will be produced as N2.
2 NO(g) N2(g) + O2(g)
1.40 M5.20 M 1.40 M
Kc =
[N2][O2] [NO]2
Kc =
[1.40][1.40] [5.20]2
= 0.0725
At 210°C, the Kc is 64.0 The equilibrium concentrations of N2 and O2 are 0.40 mol/L and 0.60 mol/L, respectively. Calculate the equilibrium concentration of NO.
1. Write out the Eq Law.
2. Rearrange for [NO].
N2(g) + O2(g) 2 NO(g)
3. Substitute concentrations then solve.
At 10.0°C, the Kc is 215.0 The equilibrium concentrations of SO2 is 9.40 mol/L. Calculate the [S] and [O2] at equilibrium.
SO2(g) S(g) + O2(g)
Values should be the same amount for S and O2 – assign the unknown “x”
x M9.40 M x M
Kc =
[S][O2] [SO2]
= Kc [S][O2] [SO2]
[x][x] = 215 [9.40]
= Kc [S][O2] [SO2]
[x]2 = 2021√ √x = 45.0 M
[S]eq and [O2] eq = 45.0 M
3. Simple I.C.E. Tables
1.00 mole of hydrogen and 1.00 mole of fluorine are sealed in a 1.00 L flask at 150.0°C and allowed to react. At equilibrium, 1.32 moles of HF are present. Calculate KC.
H2(g) + F2(g) 2 HF(g)
[Eqlbm] 1.320.340.34
[Initial] 1.00 mol/L 1.00 mol/L 0
[Change] - 0.66 - 0.66 + 1.32
1. Write out the Eq Law – for the equation as written.
[Eq] 1.320.340.34
H2(g) + F2(g) 2 HF(g)
Initially 2.0 mol of SO2, 1.0 mol of O2 are mixed in a 3.0 L reaction container. At equilibrium, 0.20 mol of O2 are found to remain. Calculate the Kc.
2 SO3 (g) 2 SO2 (g) + O2 (g)
[E] 0.15 0.0670.52
[I] 0 0.67 mol/L 0.33 mol/L
[C] - 0.52+ 0.52 - 0.26
Kc =
[SO2]2[O2] [SO3]2
Kc = 0.00557
Kc = 5.6 x 10-3
[E] 0.15 0.0670.52
Kc =
[SO2]2[O2] [SO3]2
2 SO3 (g) 2 SO2 (g) + O2 (g)
Kc =
[0.15]2[0.067] [0.52]2
4. HARD I.C.E. Tables
[E] 6.0 - x 2x6.0 - x
6.0 moles of N2 and O2 gases are placed in a 1.0 L container, what are the all concentrations at equilibrium? The Kc is 6.76.
N2(g) + O2(g) 2 NO(g)
[I] 6.0 mol/L 0 6.0 mol/L
[C] - x - x + 2x
2. Get rid of the square by taking the square root of both sides.
1. Substitute known values.
3. Isolate, and solve for x.
[I]
[C]
[E]
6.0 mol/L 0 6.0 mol/L
- 3.4 - 3.4 2(3.4)
6.82.6 2.66.0 - x 2x6.0 - x
- x - x + 2x
H2 (g) + I2 (g) 2 HI (g)
2.0 moles of HI were placed in a 1.0 L flask at 430ºC. (Kc = 54.3) Calculate the equilibrium concentrations.
[I]
[C]
[E]
0 0 2.0
+ x + x - 2x
2.0 - 2x+ x+ x
Kc = [HI]2
[H2][I2]
54.3 = [2.0 - 2x]2
[+ x][+ x]
Kc = [HI]2
[H2][I2]
7.37 = 2.0 – 2x x
7.37 x = 2.0 – 2x
√ √
7.37 x + 2x = 2.0
9.37 x = 2.0
x = 0.21
[I]
[C]
[E]
0 0 2.0
+ x + x - 2x
2.0 - 2x+ x+ x
0.21 0.21 - 0.42
1.580.210.21
The Kc for the reaction 5.3 x 10 -2 at 0ºC. Initially, 2.5 mol each particle was injected into a 1 L reaction vessel. Find the Eq concentrations.
2.5 + x 2.5 + x 2.5 - 2x-2x+ x+ x2.5 mol/L 2.5 mol/L2.5 mol/L
[E]
[C]
[I]
H2 (g) + Cl2 (g) 2 HCl (g)
Kc = [HCl]2
[H2][Cl2]
5.3 x 10 -2 = [2.5 - 2x]2
[2.5 + x][2.5 + x]
Kc = [HCl]2
[H2][Cl2]
√5.3 x 10 -2 = [2.5 - 2x]2
[2.5+x]2 √
√
0.23 = 2.5 – 2x 2.5+ x
x = 0.86 mol/L x = 1.92
2.23
0.58 + 0.23x = 2.5 - 2x
2.5 mol/L 2.5 mol/L2.5 mol/L
[E]
[C]
[I]
+ 0.86 + 0.86 2(0.86)
0.8 mol/L3.4 mol/L3.4 mol/L
2.23 x = 1.92
