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“ICE” CALCULATIONS

Learn how to calculate the equilibrium constants for reactions involving H+ and OH- ions, and apply the calculations to solve specific problems. Includes examples and step-by-step instructions.

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“ICE” CALCULATIONS

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  1. “ICE” CALCULATIONS

  2. THE EQUILIBRIUM CONSTANT aA + bB--->cC + dD (at a given T) Pure substances are not included in equation.

  3. WRITE THE EQUILIBRIUM CONSTANT EXPRESSION H+ (aq)+ OH- (aq) eH2O(l) K = CaO (s) + CO2 (g) e CaCO3 (s) K =

  4. Typical Calculations PROBLEM: Place 1.00 mol each of H2 and I2 in a 1.0 L flask. Calc. equilibrium concentrations. H2(g) + I2(g) e 2 HI(g) - - + X X e 2 X Q = 0/12 = 0

  5. H2(g) + I2(g) g 2 HI(g)Kc = 55.3 Step 1. Set up ICE table [H2] [I2] [HI] Initial 1.00 1.00 0 Change Equilib

  6. Step 2. define “X” [H2] [I2] [HI] Initial 1.00 1.00 0 Change -x -x +2x Equilib 1.00-x 1.00-x 2x where x is defined as am’t of H2 and I2 consumed on approaching equilibrium.

  7. H2(g) + I2(g) e 2 HI(g)Kc = 55.3 Step 3. Put equilibrium concentrations into Kc expression.

  8. H2(g) + I2(g) ---> 2 HI(g) Kc = 55.3 Step 4. Solve Kc expression - take square root of both sides. x = 0.79 Therefore, at equilibrium [H2] = [I2] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M

  9. Pg 132 CPSC

  10. Nitrogen Dioxide EquilibriumN2O4(g) --->2 NO2(g) X 2X e

  11. Nitrogen Dioxide EquilibriumN2O4(g) ---> 2 NO2(g) If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an ICE table [N2O4] [NO2] Initial 0.50 0 Change Equilib

  12. Nitrogen Dioxide EquilibriumN2O4(g) ---> 2 NO2(g) Step 2. Define “X” [N2O4] [NO2] Initial 0.50 0 Change -x +2x Equilib 0.50 - x 2x

  13. APPROXIMATE: K is small: x will be small Step 2. Define “X” [N2O4] [NO2] Initial 0.50 0 Change -x +2x Equilib 0.50 2x

  14. Step 3. Substitute into Kc expression and solve. Rearrange: 0.0059 (0.50) = 4x2 0.0029 = 4x2 0.0029/4 = x2 X = 0.027 Step 4. check if assumption is correct 0.50 – 0.027 = 0.47 same?

  15. Step 3. Substitute into Kc expression and solve. Rearrange: 0.0059 (0.50 - x) = 4x2 0.0029 - 0.0059x = 4x2 4x2 + 0.0059x - 0.0029 = 0 This is a QUADRATIC EQUATION ax2 + bx + c = 0 a = 4 b = 0.0059c = -0.0029

  16. Solve the quadratic equation for x. ax2 + bx + c = 0 a = 4 b = 0.0059c = -0.0029 x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027

  17. Nitrogen Dioxide EquilibriumN2O4(g) ---> 2 NO2(g) x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027 x = 0.026 or -0.028 negative value is not reasonable. Conclusion: x = 0.026 M [N2O4] = 0.050 - x = 0.47 M [NO2] = 2x = 0.052 M

  18. Solving Quadratic Equations • Recommend you solve the equation exactly on a calculator

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