Big Idea 6 – Any bond or intermolecular attraction that can be formed can be broken. These two processes are in a dynamic competition, sensitive to initial conditions and external perturbations.
•Understand the concept of dynamic equilibrium
•Be able to write an expression in terms of concentrations for the equilibrium constant Kc given a chemical equation •Understand that equilibria take a finite time to be achieved
•Be able to calculate values for Kc and associated data from initial concentrations
•Be able to write an expression in terms of partial pressures for the equilibrium constant Kp given a chemical equation •Be able to calculate values for Kp and associated data from pressure data
•Recall and understand Le Chatelier's Principle
•Understand the application of Le Chatelier's Principle and be able to predict the shift in position of equilibria and optimum conditions in reactions
•Understand and be able to apply the relationship of Kc to Kp, the different formats of Kc (reciprocals and roots) and the relationships in simultaneous equilibria
CHEMICAL EQUILIBRIUM
CHEMICAL EQUILIBRIUM: state where concentrations of products and reactants remain constant. - Equilibrium is dynamic
- Any chemical reaction in a closed vessel will reach equilibrium - At equilibrium, forward reaction RATE = reverse reaction RATE - At equilibrium, the concentrations are constant.
- Chemical equilibrium is a dynamic, or changing, process.
Previous knowledge:
Most reactions do not go all the way to completion (meaning all the reactants are used up). Reactions can proceed in both directions. This is usually indicated by a double arrow: ↔
A chemical reaction in which the products regenerate the original reactants is called a REVERSIBLE REACTION. In theory, all reactions are reversible. Some are reversible on their own and others are reversible only under restricted conditions. Do not assume that all reactions proceed until one of the reactants is entirely consumed.
* * * * * * *
The forward process: A + B → C + D
Decreases in rate, as time increases.
At equilibrium, the forward and the reverse rates become equal: A + B ↔ C + D
The reverse process: C + D → A + B
LAW OF MASS ACTION
aA + bB ↔ cC + dD K = [𝐶]𝑐[𝐷]𝑑
[𝐴]𝑎[𝐵]𝑏
- Remember: products divided by reactants
- [ ] stands for the concentration – mol/liter, call MOLARITY, M
- Kc = K = Keq = equilibrium constant. These are used interchangeably. - Example: 4NH3(g) + 7O2(g) ↔ 4NO2(g) + 6H2O(g) K = [NO2]4[H2O]6
[NH3]4[O2]7
- K is a mathematical expression that describes the equilibrium state associated with a chemical change.
- K changes with temperature but NOT with pressure and concentration. - The units for K depend on the reaction. They are not usually used.
PRACTICE:
Write the rate expressions for the following reactions:
1. 3H2(g) + N2(g) 2NH3(g) 2. SO2(g) + NO2(g) NO(g) + SO3(g)
* * * * * * *
For the REVERSE REACTION: cC + dD ↔ aA + bB K′ = [𝐴]𝑎[𝐵]𝑏
[𝐶]𝑐[𝐷]𝑑 K = 1/K = K´
If the original reaction is multiplied by some factor, n, to give naA + nbB ↔ ncC + ndD, then K′′ = [𝐴]𝑛𝑎[𝐵]𝑛𝑏
[𝐶]𝑛𝑐[𝐷]𝑛𝑑 K´´ = K
n
PRACTICE:
At a given temperature, K = 1.3 x 10-2 for the reaction N2(g) + 3H2(g) ↔ 2NH3(g). Calculate the values of K for the following reactions at this temperature.
a. 1
2N2(g) + 3
2H2(g) ↔ NH3(g).
b. 2NH3(g) ↔ N2(g) + 3H2(g)
c. NH3(g) ↔ 1
2N2(g) + 3
d. 2N2(g) + 6H2(g) ↔ 4NH3(g).
CALCULATING A VALUE FOR K
STEPS:
1. Write the balanced chemical equation.
2. Write down the concentrations for all reactants and products. Use the equilibrium
concentrations.
3. Use the relationship to solve the problem
EXAMPLE:
What is the equilibrium constant if the final concentrations are NOCl = [0.625], NO = [0.220], and Cl2 = [0.337]. NOCl gas decomposes to form NO gas and Cl2 gas.
1. 2NOCl(g) ↔ 2NO(g) + Cl2(g)
2. and 3. K = [Products] = [NO]2[Cl2] = [0.220]2 [0.337] = 0.420 [Reactants] [NOCl]2 [0.625]2
HETEROGENEOUS EQUILIBRIA
Many equilibria comprise of more than one phase and are called HETEROGENEOUS EQUILIBRIA. Experimental results show that the position of a heterogeneous equilibrium does not depend on the amounts of pure solids and liquids present. The concentrations of these type of substances does not change. Therefore, do not include them in the mass action expression.
PRACTICE:
a. 3Fe(s) + 2H2O(g) ↔ Fe3O4(s) + 2H2(g) b. Al(s) + H2O(g) ↔ Al2O3(aq) + H2(g)
SOME THINGS TO REMEMBER:
1. Can only use gases and aqueous solutions in the rate expressions. Do not include pure solids or pure liquids in your mass action expression. The concentrations of pure solids and liquids are constants that can be determined from their densities. If it is an aqueous solution, it is not pure, and must included.
2. The EQUILIBRIUM POSITION (not constant) is a set of equilibrium concentrations. It
depends on the initial concentration while K does not. There is only on K for a given reaction at a given temperature, but there are a infinite number of equilibrium positions.
4. If K is large (greater than one), then products are favored and equilibrium is established after lots of product is formed.
5. If K = 1, the concentrations of the reactants and products are equal.
6. Each reaction has a unique Keq for every temperature. If the temperature changes, the Keq changes. 7. Keq does not indicate the time it takes to reach equilibrium. It only provides information about the
mixture of reactants and products at equilibrium.
8. K values cannot always be directly compared because the stoichiometry differs.
EQUILIBRIUM EXPRESSIONS INVOLING PRESSURES
Pressures can be used in equilibrium expressions. The equilibrium constant is called Kp. Using the same mass action equation as before, the K expression becomes Kp:
K = [𝐶] 𝑐[𝐷]𝑑
[𝐴]𝑎[𝐵]𝑏 becomes K𝑝 =
[𝑃𝑐]𝑐[𝑃𝑑]𝑑 [𝑃𝑎]𝑎[𝑃𝑏]𝑏
P = partial pressure at equilibrium in atmospheres, atm.
Kc or K involve concentrations while Kp involves pressures. Kp and K can be interconverted using the following relationship:
Kp = K(RT)∆n R = 0.0821 L∙atm/mol∙K T = kelvin temperature
n = #moles gaseous product - # moles of gaseous reactant
K = Kp if # of moles gaseous product = # moles gaseous reactant.
PRACTICE:
Consider the reaction 2NOCl(g) ↔ 2NO(g) + Cl2(g) at 35°C where 3.00mol NOCl, 1.00mol NO, and 2.00 Cl2 are mixed in a 10.0L flask After the system has reached EQ, the concentrations are observed to be [Cl2] = 1.52 x 10-1M, [NO] = 4.00 x 10-3M, and [NOCl] = 3.96 x 10-1M.
a. Calculate the value of K for this systems at 35°C.
b. Calculate the value of K for the reverse reaction.
d. Calculate the value of K for the reaction 4NOCl(g) ↔ 4NO(g) + 2Cl2(g)
THE REACTION QUOTIENT, Q
It is not always obvious if a mixture has reached equilibrium or not. If it is not at equilibrium, then it is helpful to know which direction the reaction must proceed in order to reach equilibrium. To determine this, solve for the REACTION QUOTIENT, Q. Q only needs to be calculated when there is SOME OF EACH REACTANT AND PRODUCT PRESENT.
The reaction quotient allows you to calculate the problem and then compare it to K. The reaction quotient can be calculated by substituting these concentrations into the equilibrium expression.
Deciding Direction – always compare Q to K:
• If Q is less than K, this means that at the time of measurement, there were too much of the reactants and too little of the products. The reaction will proceed to the right, in the direction of the products.
• If Q is more than K, this means that at the time of measurement, there were too little of the reactants and too much of the products. The reaction will proceed to the left, in the direction of the reactants.
• If Q is equal to K, the system is at equilibrium and there will be no shift in direction at this temperature.
PRACTICE:
For the synthesis of ammonia, the value of K is 6 x 10-2 at 500°C. In an experiment, 0.50mol N2, 1.0 x 10-2mol H2, and 1.0 x 10-4mol NH3 are mixed at 500°C in a 1.0L flask. In which direction will the system proceed to equilibrium? N2(g) + 3H2(g) ↔ 2NH3(g).
RICE or ICE PROBLEMS
STEPS FOR SOLVING EQ PROBLEMS
1. Write the balanced chemical equation. If a chemical reaction occurs, work out the
STOICHIMETRY and then write a second equation for the EQUILIBRIUM reaction. Always do stoichiometry in moles first.
2. Set up the equilibrium expression (no numbers yet)
3. If you cannot tell which way the reaction is going to shift, solve for Q.
5. Substitute these final concentrations into the equilibrium expression and solve for X. 6. Check your answer to make sure that it is logical.
When solving an EQ problem, some +x or –x values can be sometimes be treated as negligible. It is considered negligible if it is less than 5% of the number from which is was subtracted. If x is NOT negligible, the quadratic equation must be used.
EXAMPLE
At 700K, carbon monoxide reacts with water to form carbon dioxide and hydrogen: CO(g) + H2O(g) ↔ CO2(g) + H2(g). The equilibrium constant for this reaction is 5.10. Consider and experiment in which 1.00mol CO and 1.00mol H2O are mixed together in a 1.00L flask at 700K. Calculate the
concentrations of all species at equilibrium.
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
Initial 1.00M 1.00M 0 0
Change -x -x +x +x
EQ 1.00-x 1.00-x x x
K = [𝐶𝑂2] [𝐻2]
[𝐶𝑂] [𝐻2𝑂]
= 5.10 =
𝑥2 (1.00−𝑥)2
𝑥
1.00−𝑥
= √5.10 = 2.26
x = 2.26(1.00-x) = 2.26 – 2.26x 2.26 = x + 2.26x = 3.26x
x = 2.26
3.26 = 0.693M [H2O] = [CO] = 1.00-0.693 = 0.307 = 0.31M
[CO2] = [H2] = 0.693M
PRACTICE
2. Consider the reaction 2HF(g) ↔ H2(g) + F2(g), where K = 1.0 x 10-2 at some very high temperature. In an experiment, 5.00mol of HF, 0.500mol H2, and 0.750mol F2 are mixed in a 5.00L flask and allowed to react to equilibrium. Solve for the equilibrium concentrations. a. Solve for molarity for each.
b. Solve for Q – you have reactants and products, so you must. This will let you know which variables will be positive and which will be negative.
c. Do the RICE calculation. If it is NOT a perfect square, assume that the x value is negligible since K is very small. Can only make this assumption when K is very small.
LE CHATELIER’S PRINCIPLE
If a change in conditions is imposed on a system at equilibrium, the equilibrium position will shift in the direction that tends to reduce that change in conditions. A stress is any kind of change in a system at equilibrium that upsets the equilibrium. A reaction system will shift in the forward or reverse direction to “undo” the altering factor.
CHANGES IN CONCENTRATION
An increase in concentration of a reactant will cause equilibrium to shift to the right to form more products. An increase in concentration of the product will cause a shift to the left to make more reactants.
A decrease in concentration of a product will cause equilibrium to shift to the right to form more products. A decrease in concentration of the reactant will cause a shift to the left to make more reactants.
The position of equilibrium shifts, the equilibrium constant does not. If the reaction quotient, Q, is greater than Keq, the reaction will shift to the left in order to reach equilibrium. If Q is less than Keq, the reaction will shift to the right to return to equilibrium.
CHANGES IN PRESSURE (THEREFORE VOLUME)
Changes in pressure only affect equilibrium systems having gaseous products and/or reactants.
increasing the pressure of a gaseous system will cause equilibrium to shift to the side with fewer gas particles. decreasing the pressure of a gaseous system will cause equilibrium to shift to the side with
more gas particles. If the system has the same number of moles on both sides, changes in pressure does
not affect equilibrium.
CHANGING THE TEMPERATURE
Remember that the value of the equilibrium constant for a particular reaction depends on temperature.
An increase in temperature of an exothermic reaction will cause equilibrium to shift to the left. A decrease in temperature of an exothermic reaction will cause equilibrium to shift to the right.
An increase in temperature of an endothermic reaction will cause equilibrium to shift to the right. A
decrease in temperature of an endothermic reaction will cause equilibrium to shift to the left.
ADDITION OF A CATALYST
Adding a catalyst does not affect equilibrium. It only affects the rate of a reaction.
PRACTICE:
Consider the reaction 2NO2(g) ↔ N2(g) + 2O2(g) which is exothermic. A vessel contains NO2(g), N2(g), and O2(g) at equilibrium. Predict how each of the following stresses will affect the concentration of O2 and the value of K.
[O2] K
A. NO2 is added B. N2 is removed
C. The volume is halved D. He(g) is added