CH 117 Spring 2015Worksheet 6

  1. For the reaction PCl3 (g) + Cl2 (g) ⇌ PCl5 (g), the value of K = 96.2 at 400 K. If the initial concentrations are 0.22 M of both PCl3 and Cl2, what are the equilibrium concentrations of all species?

Set up an ICE chart (always the method for questions asking for equilibrium concentrations!)

PCl3 / Cl2 / PCl5
Initial / 0.22 / 0.22 / 0
Change / -x / -x / +x
Equilibrium / 0.22-x / 0.22-x / x

Plug into K expression and solve for x.

K = 96.2 = [PCl5]/[Cl2][PCl3] = x/(0.22 –x)(0.22 – x) = x/(0.22 – x)2

Multiply both sides by denominator to get isolate x.

96.2(0.22 – x)2 = x  expand left side  96.2(0.0484 – 0.44x + x2) = x  continue expanding  4.65608 – 42.328x + 96.2x2 = x  rearrange into quadratic form

4.65608 – 43.328x + 96.2x2 = 0  solve for x using the quadratic formula

x = 0.2733 or x = 0.1771  when you get 2 positive values for x, plug both into the equilibrium row and solve for concentrations. One of the values will give you a negative concentration, which is impossible, so you can choose the other value of x.

If x = 0.2733  [PCl3] = 0.22 – 0.2733 = -0.0533  not possible for this to be the value of x, so we know it must be the other value and x = 0.1771.

[PCl3] = [Cl2] = 0.22 – 0.1771 = 0.0429 M

[PCl5] = 0.1771 M

  1. For the reaction N2 (g) + C2H2 (g) ⇌ 2 HCN (g), K = 2.3 x 10-4 at 300 K. What is the equilibrium concentration of HCN if the initial concentrations of N2 and C2H2 are 3.5 M and 2.5 M respectively?

Set up an ICE chart.

N2 / C2H2 / 2 HCN
Initial / 3.5 / 2.5 / 0
Change / -x / -x / +2x
Equilibrium / 3.5 – x / 2.5 – x / 2x

Plug into K expression and solve for x.

K = 2.3 x 10-4 = [HCN]2/[N2][C2H2] = (2x)2/(3.5 – x)(2.5 –x) = 4x2/(8.75 – 6x + x2)

Multiply both sides by denominator.

2.3 x 10-4(8.75 – 6x + x2) = 4x2 Distribute 

0.0020125 – 0.00138x + 0.00023x2 = 4x2  rearrange to get into quadratic form and solve for x using the quadratic formula.

0.0020125 – 0.00138x – 3.99977x2 = 0  x = -0.0226 or x = 0.0223  choose positive x value and plug it in to equilibrium row to solve for concentrations.

[N2] = 3.5 – 0.0223 = 3.48 M

[C2H2] = 2.5 – 0.0223 = 2.48 M

[HCN] = 2(0.0223) = 0.0446 M

  1. Describe all of the components of Le Chatelier’s Principle.

Le Chatelier’s principle is all about getting a system back to equilibrium once the equilibrium has been disturbed. Le Chatelier looked at factors that would change equilibrium and predicted how they would change it. The “three” factors he looked at were concentration, pressure/volume, and temperature.

Concentration – Think of changing concentration like a seesaw. If you add more of a reactant, the reaction will shift to get rid of those reactants and make more products. If you add more products, the reactant will shift to make more reactants. However, if you remove a reactant, the reaction will shift to make reactants and replace the loss. If you remove a product, the reaction will shift to make products.

Pressure/Volume – Pressure and volume are inversely related (as pressure goes up, volume goes down and vice versa). For that reason, we can group them together when we discuss equilibrium. What is true for an increase in pressure, would be the same for a decrease in volume, etc. The effects of pressure and volume have to do with the number of moles of gas on each side of the reaction. For that reason, if there is the same number of moles of gaseous reactants as there are gaseous products, pressure and volume WILL NOT affect equilibrium. If you decrease volume (increase pressure), the reaction will shift to the side with less moles of gas. If you increase volume (decrease pressure), the reaction will shift to the side with more moles of gas.

Temperature –A change in temperature depends on the enthalpy of a reaction (whether it is exothermic or endothermic). An endothermic reaction has a positive heat of reaction, while an exothermic reaction has a negative heat of reaction. The easiest way to visualize a change in temperature is to place the heat in the reaction – heat will be a reactant in an endothermic reaction and a product in an exothermic reaction. A change in temperature can also affect the value of the equilibrium constant K. These changes are summarized in the table below.

Endothermic / Exothermic
Increase temperature (add more heat) / Shifts to products, increases value of K / Shifts to reactants, decreases value of K
Decrease temperature (take away heat) / Shifts to reactants, decreases value of K / Shifts to products, increases value of K
  1. In which direction would the reaction 3 Fe (s) + 4 H2O (g) ⇌ Fe3O4 (s) + 4 H2 (g) [H = -89.9 kJ] shift and how would the value of Keq change for each of the following scenarios?
  1. Addition of Fe (s)– no change, solids don’t affect equilibrium
  2. Removal of H2O (g) – shift left
  3. Increasing the concentration of Fe3O4 (s)– no change
  4. Decreasing the concentration of H2 (g) – shift right
  5. Increasing the pressure of the system – no change (same # of moles of gas on both sides)
  6. Increasing the volume of the system – no change
  7. Increasing the temperature of the system – shifts right, increases K (K ONLY changes with temperature!)