1. Quantitative Changes in Equilibrium Systems
Lesson 4
7.5

2. Reaction Quotient (Q)
The reaction quotient, Q, is an expression that is identical to the equilibrium constant expression, but its value is calculated using concentrations that are not exactly at equilibrium.

3. Reaction Quotient (Q) Consider:
If Q = K the system must be at equilibrium
If Q › K the numerator must be very large ([products/right side] greater than at equilibrium) and the system must attain equilibrium by moving to the left
If Q ‹ K the denominator must be very large ([reactants/left side] greater than at equilibrium) and system must attain equilibrium by moving to the right

4. N2(g) + 3H2(g)  2NH3(g) ∆H = -92kJ/mol
Determining the Direction Shift to Attain Equilibrium Example Problem #1
N2(g) + 3H2(g)  2NH3(g) ∆H = -92kJ/mol
At 500oC, the value of K for the reaction is The following concentrations of gases are present in a container at 500oC:
[N2(g)] = 0.10 mol/L [H2(g)] = 0.30 mol/L [NH3(g)] = 0.20 mol/L
Is this mixture of gases at equilibrium? If not, in which direction will the reaction go to reach equilibrium?

5. EQUILIBRIUM CALCULATIONS
organize data with an ICE table (initial, change, equilibrium)
letting x represent the substance with the smallest coefficient in the chemical equation avoids fractional values of x
look for perfect squares when solving an equilibrium expression (only works if square on top and bottom)
take the square root of each side to solve the equation
remember that the square root is ± x
many problems do not involve perfect squares
you may need to use the quadratic equation to solve the expression
Recall that a quadratic equation of the form:
In order to solve use:

6. IGNORING AN x VALUE
In general, a difference of less than 5% justifies the simplifying assumption. If can be shown that, if the concentration to which x is added or subtracted is at least times greater than the value of K, the simplifying assumption will give you an error of less than 5%.

7. Known as the 500 rule
The change in the initial concentration, x, is negligible and can be ignored.
The change in the initial concentration, x, may not be negligible. The equilibrium equation will be more complex, possibly requiring the solution of a quadratic equation.
*[A] = initial concentration

8. Calculating an Equilibrium Constant By Taking the Square Root
The following reaction increases the proportion of hydrogen gas for use as a fuel.
CO(g) + H2O(g)  H2(g) + CO2(g)
This reaction had been studied at different temperatures to find the optimum conditions. At 700.0K, the equilibrium constant is 83. Suppose that you start with 1.0mol of CO(g)? and 1.0mol H2O(g) in a 5.OL container. What amount of each substance will be present in the container when the gases are at equilibrium at 700.0k?

9. Calculating an Equilibrium Constant By Taking the Square Root
The following reaction increases the proportion of hydrogen gas for use as a fuel.
CO(g) + H2O(g)  H2(g) + CO2(g)
This reaction had been studied at different temperatures to find the optimum conditions. At 700.0K, the equilibrium constant is 83. Suppose that you start with 1.0mol of CO(g)? and 1.0mol H2O(g) in a 5.OL container. What amount of each substance will be present in the container when the gases are at equilibrium at 700.0k?

10. Calculating an Equilibrium Constant By Assuming x is negligible
Carbon monoxide gas, CO(g), is a primary starting material in the synthesis of many organic compounds, including methanol CH3OH(l). At °C, K is 6.40 x 10-7 for the decomposition of carbon dioxide gas, CO2(g), into carbon monoxide and oxygen, O2(g). Calculate the concentrations of all entities at equilibrium if mol of CO2(g) is placed in a 1.000L closed container and heated to °C.

11. Calculating an Equilibrium Constant By Assuming x is negligible
Carbon monoxide gas, CO(g), is a primary starting material in the synthesis of many organic compounds, including methanol CH3OH(l). At 2000 °C, K is 6.40 x 10-7 for the decomposition of carbon dioxide gas, CO2(g), into carbon monoxide and oxygen, O2(g). Calculate the concentrations of all entities at equilibrium if mol of CO2(g) is placed in a 1.000L closed container and heated to 2000°C.

12. Calculating an Equilibrium Constant By Using The Quadratic Formula
Suppose that hydrogen fluoride gas, HF(g), was synthesized by combining 3.00mol of hydrogen gas, H2(g) and mol of fluorine gas, F2(g) in a 3.00L sealed flask. Assume that the equilibrium constant for the synthesis reaction at this temperature is 1.15 x 102. What are the concentrations for each gas at equilibrium?

13. Calculating an Equilibrium Constant By Using The Quadratic Formula
Suppose that hydrogen fluoride gas, HF(g), was synthesized by combining 3.00mol of hydrogen gas, H2(g) and mol of fluorine gas, F2(g) in a 3.00L sealed flask. Assume that the equilibrium constant for the synthesis reaction at this temperature is 1.15 x 102. What are the concentrations for each gas at equilibrium?

14. Calculating an Equilibrium Constant By Using The Quadratic Formula
Suppose that hydrogen fluoride gas, HF(g), was synthesized by combining 3.00mol of hydrogen gas, H2(g) and mol of fluorine gas, F2(g) in a 3.00L sealed flask. Assume that the equilibrium constant for the synthesis reaction at this temperature is 1.15 x 102. What are the concentrations for each gas at equilibrium?

15. Homework Answer Practice Problems Pg. 452 #1-3, Pg. 454 #1-3,