Quantitative Changes in Equilibrium Systems LESSON 4 7

Quantitative Changes in Equilibrium Systems LESSON 4 7. 5

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.

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

Determining the Direction Shift to Attain Equilibrium Example Problem #1 N 2(g) + 3 H 2(g) 2 NH 3(g) ∆H = -92 k. J/mol At 500 o. C, the value of K for the reaction is 0. 40. The following concentrations of gases are present in a container at 500 o. C: [N 2(g)] = 0. 10 mol/L [H 2(g)] = 0. 30 mol/L [NH 3(g)] = 0. 20 mol/L Is this mixture of gases at equilibrium? If not, in which direction will the reaction go to reach equilibrium?

EQUILIBRIUM CALCULATIONS 1. organize data with an ICE table (initial, change, equilibrium) 2. letting x represent the substance with the smallest coefficient in the chemical equation avoids fractional values of x 3. 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 4. 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:

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 500 times greater than the value of K, the simplifying assumption will give you an error of less than 5%.

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

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) + H 2 O(g) H 2(g) + CO 2(g) This reaction had been studied at different temperatures to find the optimum conditions. At 700. 0 K, the equilibrium constant is 83. Suppose that you start with 1. 0 mol of CO(g)? and 1. 0 mol H 2 O(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. 0 k?

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) + H 2 O(g) H 2(g) + CO 2(g) This reaction had been studied at different temperatures to find the optimum conditions. At 700. 0 K, the equilibrium constant is 83. Suppose that you start with 1. 0 mol of CO(g)? and 1. 0 mol H 2 O(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. 0 k?

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 CH 3 OH(l). At 2000. 0 °C, K is 6. 40 x 10 -7 for the decomposition of carbon dioxide gas, CO 2(g), into carbon monoxide and oxygen, O 2(g). Calculate the concentrations of all entities at equilibrium if 0. 250 mol of CO 2(g) is placed in a 1. 000 L closed container and heated to 2000. 0°C.

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 CH 3 OH(l). At 2000 °C, K is 6. 40 x 10 -7 for the decomposition of carbon dioxide gas, CO 2(g), into carbon monoxide and oxygen, O 2(g). Calculate the concentrations of all entities at equilibrium if 0. 250 mol of CO 2(g) is placed in a 1. 000 L closed container and heated to 2000°C.

Calculating an Equilibrium Constant By Using The Quadratic Formula Suppose that hydrogen fluoride gas, HF(g), was synthesized by combining 3. 00 mol of hydrogen gas, H 2(g) and 6. 00 mol of fluorine gas, F 2(g) in a 3. 00 L 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?

Calculating an Equilibrium Constant By Using The Quadratic Formula Suppose that hydrogen fluoride gas, HF(g), was synthesized by combining 3. 00 mol of hydrogen gas, H 2(g) and 6. 00 mol of fluorine gas, F 2(g) in a 3. 00 L 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?

Calculating an Equilibrium Constant By Using The Quadratic Formula Suppose that hydrogen fluoride gas, HF(g), was synthesized by combining 3. 00 mol of hydrogen gas, H 2(g) and 6. 00 mol of fluorine gas, F 2(g) in a 3. 00 L 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?

Homework Answer Practice Problems Pg. 452 #1 -3, Pg. 454 #1 -3, Pg. 458 #1 -3, Pg. 459 #1 -8