Use “ ICE ” to help you find the concentrations of each substance at equilibrium. I nitial C hange E quilibrium.

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ICE Tables I nitial (Concentrations) C hange E quilibrium (Concentrations)

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Presentation transcript:

Use “ ICE ” to help you find the concentrations of each substance at equilibrium. I nitial C hange E quilibrium

1. When 0.40 moles of PCl 5 is heated in a 10.0 L container, an equilibrium is established in which 0.25 moles of Cl 2 is present. a) What is the number of moles of PCl 5 and PCl 3 at equilibrium? b) What are the equilibrium concentration of all three components?

a) PCl 5 PCl 3 + Cl 2 Initial 0.40 mol Change  0.25 mol mol mol Equil mol 0.25 mol 0.25 mol b) [PCl 5 ] = 0.15 moles = moles/L 10.0 L [PCl 3 ] = [Cl 2 ] = 0.25 moles = moles/L 10.0 L

2. The equilibrium constant for the reaction represented below is 50 at 448°C H 2(g) + I 2(g) 2 HI (g) a) How many moles of HI are present at equilibrium when 1.0 moles of H 2 is mixed with 1.0 moles of I 2 in a 0.50 L container and allowed to react at 448°C? b) How many moles of H 2 and I 2 are left unreacted? c) If the conversion of H 2 and I 2 to HI is essentially complete, how many moles of HI would be present? d) What is the percent yield of the equilibrium mixture?

a) The concentrations at the start are: [H 2 ] = 1.0 mol/0.50 L = 2.0 mol/L [I 2 ] = 1.0 mol/0.50 L = 2.0 mol/L [HI] = 0 moles therefore 0 mol/L

Fill in the numbers under the reaction for the starting concentrations. Let 'x' = number of moles of H 2 consumed per litre H 2 + I 2 2 HI I 2.0 M 2.0 M 0 M C  x  x 2(+x) E 2.0  x 2.0  x 2x

7.1 = 2x x 50 = (2x) x = 2x (2.0 - x)(2.0 - x)x = 1.56 mol/L 50 = (2x) 2. (2.0 - x) 2 You are solving for 'x' which is in moles/L (M) therefore you can automatically fill in the units for any value of 'x'.

[H 2 ] = 2.0 M  x = 2.0 M  1.56 M = 0.44 M [I 2 ] = 2.0 M  x = 2.0 M  1.56 M = 0.44 M [HI] = 2x = M = 3.12 M

b) From part a) you get the M of H 2 and I 2 left unreacted. This is the concentration not the answer to the question. You are asked for the number of moles left unreacted therefore moles = concentration volume [H 2 ] = [I 2 ] = 0.44 mol/L 0.5 L= 0.22 moles Therefore 0.22 moles of H 2 and 0.22 moles of I 2 are left unreacted.

c) If you look at the equation again you can see that there is exactly the right amount of H 2 to react with I 2. Theoretically if both the H 2 and I 2 where all used up then we should make 2 moles of HI. This is the theoretical yield because it is what could be made in theory if everything went to completion.

d) The percentage yield is calculated by using the actual yield from part a) and the theoretical yield from part b). You may use either moles or moles/L but make sure that they are both the same. Percentage Yield = Actual Yield 100 Theoretical Yield = 1.56 M M = 78% yield (note that the moles/L units cancel)

3. How many moles of HI are present at equilibrium when 2.0 moles of H 2 is mixed with 1.0 moles of I 2 in a 0.50 L container and allowed to react at 448°C. At this temperature K eq = 50. H 2(g) + I 2(g) 2 HI (g) The concentrations at the start are: [H 2 ] = 2.0 mol/0.50 L = 4.0 mol/L [I 2 ] = 1.0 mol/0.50 L = 2.0 mol/L [HI] = 0 moles therefore 0 mol/L

Let 'x' = number of moles of H 2 consumed per litre H 2 + I 2 2 HI I 4.0 M 2.0 M 0 M C  x  x 2(+x) E 4.0  x 2.0  x 2x

50 = (2x) 2. (4.0 - x)(2.0 - x) 50 = 4x x + x x + 50x 2 = 4x 2 46x x = 0 (standard form) Use the quadratic equation to solve for 'x'.

In this case a = +46, b = -300 and c = We only started with 4 mols/L of H 2 to begin with, so 4.7 is to large. It is therefore the unreal root. So let x = 1.9 mol/L.

[H 2 ] = 4.0 M  x = 4.0 M  1.9 M = 2.1 M [I 2 ] = 2.0 M  x = 2.0 M  1.9 M = 0.1 M [HI] = 2x = M = 3.8 M The moles of HI present at equilibrium would be: moles = concentration volume = 3.8 moles/L 0.5 L = 1.9 moles