 ADVANCED PLACEMENT CHEMISTRY EQUILIBRIUM. Chemical equilibrium * state where concentrations of products and reactants remain constant *equilibrium is.

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ADVANCED PLACEMENT CHEMISTRY 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

Law of Mass Action for jA + kB  lC + mD [ ] = concentration in mol/L = Molarity

K c = K = K eq equilibrium constant (used interchangeably)

Ex. 4 NH 3 (g) + 7 O 2 (g)  4 NO 2 (g) + 6 H 2 O(g) If we know equilibrium concentrations, we can calculate the equilibrium constant, K c.

K changes with temperature (not with concentration or pressure).

For the reverse reaction: lC + mD  jA + kB K c ' = 1/ K c

If the original reaction is multiplied by some factor to give: njA + nkB  nlC + nmD

For a 2-step reaction with K c1 and K c2 as K c values for each step, the K c for the overall reaction is K c3 = K c1 × K c2.

The units for K depend on the reaction. They are usually not used.

Equilibrium position - a set of equilibrium concentrations. - depends on initial concentrations. (K c doesn't)

Pressures can be used in equilibrium expressions. The equilibrium constant is called K p. Using the same mass action equation as above, the K p expression becomes: P = partial pressure at equilibrium in atm. Don’t use brackets! (you may see this without parenthesis)

K c involves concentrations while K p involves pressures. K p and K c can be interconverted using the following relationship : K p = K c (RT)  n R = 0.08206 Latm/molK T = Kelvin temperature  n = #moles gaseous product - # moles gaseous reactant K c = K p if # moles gaseous product = # moles gaseous reactant

***Concentrations of pure solids and pure liquids are not included in equilibrium expressions because they are constant. CaCO 3 (s)  CaO (s) + CO 2 (g) K c = [CO 2 ]

*A value for K greater than one means that the equilibrium is far to the right (mostly products). *A value for K less than one means that the equilibrium is far to the left (mostly reactants). *The size of K and the time needed to reach equilibrium are not directly related. *K values can not always be directly compared because stoichiometry differs.

If the value for K is very small, reactants are present in great excess at equilibrium. If the value for K is very large, products are present in great excess at equilibrium. Values for K in the range of 0.001 to 1000 describe reactions where reactants and products are both present in significant quantities at equilibrium. Remember that a large K does not necessarily mean the reaction is fast!

Consider the reaction 2NOCl(g)  2NO(g) + Cl 2 (g) at 35 o C. Various amounts of NOCl(g), NO(g), and Cl 2 (g) are mixed in a 10.0 L flask. After the system has reached equilibrium the concentrations are observed to be: [Cl 2 ] = 1.52 x 10 -1 M [NO] = 4.00 x 10 -3 M [NOCl] = 3.96 x 10 -1 M Calculate the value of K for this system at 35 o C. = 1.55 × 10 -5 Always set up the expression w/o numbers first! Never use initial concentrations, only equilibrium concentrations!

Calculate the value of K for the reaction 2NO(g) + Cl 2 (g)  2NOCl(g). K' = 1/K K' = 1/1.55×10  5 = 6.45×10 4

Calculate the value of K for the reaction 4NOCl(g)  4NO(g) + 2Cl 2 (g). K" = K n K" = (1.55×10  5 ) 2 = 2.40×10  10

Calculate K p for the first reaction. K p = K c (RT)  n K p = (1.55×10  5 )[(0.0821)(308)] 1 = 3.92×10  4

When working equilibrium problems, it is not always obvious which direction that equilibrium is going to shift. *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.

Q = reaction quotient - calculated like K c, but uses initial concentrations instead of equilibrium concentrations 1. If K = Q, then system is at equilibrium (no shift occurs) 2. If Q > K, [product]/[reactant] is too large (system shifts to left) 3. If Q < K, [product]/[reactant] is too small (system shifts to the right)

Example 2. For the synthesis of ammonia, the value of K is 6×10  2 at 500 o C. In an experiment, 0.50 mol of N 2 (g), 1.0×10  2 mol of H 2 (g), and 1.0×10  4 mol of NH 3 (g) are mixed at 500 o C in a 1.0 L flask. In which direction will the system proceed to reach equilibrium? N 2 + 3 H 2  2NH 3 Initial 0.50 mol/L 0.010 mol/L 1.0 x 10 -4 mol/L Q = [NH 3 ] 2 = (1.0×10  4 ) 2 = 2.0×10  2 [N 2 ][H 2 ] 3 (0.50)(1.0×10  2 ) 3 Since Q < K (2.0×10  2 < 6×10  2 ), the reaction will shift to the right to reach equilibrium.

When solving equilibrium problems, it is very important to follow a series of steps. Skipping these can lead to problems (and fewer points on your AP exam).

STEPS FOR SOLVING EQUILIBRIUM PROBLEMS 1. Write a balanced equation. If a chemical reaction occurs, work out the stoichiometry 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!) These steps will be used for the next 4 chapters!!!!

3. If you can't tell which way the reaction is going to shift, solve for Q. 4. Set up a chart that includes the equation, initial concentrations, changes in concentration in terms of x, and final concentrations.

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 equilibrium problem, some +x and -x values can be treated as negligible. x is considered negligible if it is less that 5% of the number that it was to be subtracted from or added to. If x is not negligible, the quadratic equation must be used. On the AP test, problems will not require the quadratic equation.

At 700 K, carbon monoxide reacts with water to form CO 2 and H 2 : CO(g) + H 2 O(g)  CO 2 (g) + H 2 (g) The equilibrium constant for this reaction at 700 K is 5.10. Consider an experiment in which 1.00 mol of CO(g) and 1.00 mol of H 2 O(g) are mixed together in a 1.00 L flask at 700 K. Calculate the concentrations of all species at equilibrium. Reaction CO + H 2 O  CO 2 + H 2 Initial 1.00 M 1.00 M 0 0 Change  x  x +x +x Equil. 1.00  x 1.00  x x x  x = 0.69M [H 2 O]=[CO]= 1.00-0.69 = 0.31 M [CO 2 ]=[H 2 ]= 0.69 M This is a perfect square. Solve by taking the square root of both sides.

Example 4. Calculate the number of moles of Cl 2 produced at equilibrium in a 10.0 L vessel when 1.00 mol of PCl 5 is heated to 250 o C. K = 0.041 mol/L 1.00 mol/10.0 L = 0.100 M R PCl 5 (g)  PCl 3 (g) + Cl 2 (g) I 0.100 0 0 C  x +x +x E 0.100  x x x K c = [PCl 3 ][Cl 2 ] 0.041 = x 2  x 2 [PCl 5 ] 0.100  x 0.100 x = 0.064M 0.064 is more than 5% of 0.100 so this is not a valid approximation. This is not a perfect square. Try to solve by assuming that the –x is negligible.

Use of the quadratic equation: x 2 = 0.0041  0.041x x 2 + 0.041x – 0.0041 = 0 x = 0.047 and –0.088(not possible) [Cl 2 ] = 0.047 moles Cl 2 = 0.047M × 10.0 L = 0.47 moles Cl 2

Example 5. Consider the reaction 2HF(g)  H 2 (g) + F 2 (g) where K = 1.0×10  2 at some very high temperature. In an experiment, 5.00 mol of HF(g), 0.500 mol of H 2 (g), and 0.750 mol of F 2 (g) are mixed in a 5.00 L flask and allowed to react to equilibrium. Will the concentrations of the products increase, decrease, or remain the same when equilibrium is reached? 5.00 mol/5 L = 1.00 M HF 0.500 mol/5 L = 0.100 M H 2 0.750 mol/5 L = 0.150 M F 2 Q = [H 2 ][F 2 ] =(0.100)(0.150) = 0.015 [HF] 2 (1.00) Q > K so reaction shifts left Concentrations of products will decrease. Since we have some of each reactant and product, we have to solve for Q to determine which way the reaction will shift.

2 H 2 S(g)  2 H 2 (g) + S 2 (g) When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of H 2 S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72×10 –2 mol of S 2 (g) is present at equilibrium. a)Write the expression for the equilibrium constant, K c, for the decomposition reaction represented above. K c = [H 2 ] 2 [S 2 ] [H 2 S] 2

2 H 2 S(g)  2 H 2 (g) + S 2 (g) When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of H 2 S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72×10 –2 mol of S 2 (g) is present at equilibrium. (b) Calculate the equilibrium concentration, in mol  L -1, of the following gases in the container at 483 K. (i)H 2 (g) (ii)H 2 S(g) 3.40g H 2 S × 1 mol H 2 S = 0.0798 M H 2 S 1.25 L 34.08g H 2 S 3.72 × 10 -2 mol / 1.25 L = 0.0298 M S 2 R 2 H 2 S(g)  2 H 2 (g) + S 2 (g) I 0.0798 M 0 0 C -2x +2x +x E 0.0798-2x 2x x x = 0.0298 M (i) [H 2 ] = 2 × 0.0298 = 0.0596M [H 2 S] = 0.0798-0.0596 = 0.0202 M

2 H 2 S(g)  2 H 2 (g) + S 2 (g) When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of H 2 S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72×10 –2 mol of S 2 (g) is present at equilibrium. (c)Calculate the value of the equilibrium constant, K c, for the decomposition reaction at 483 K. K c = (0.0596) 2 (0.0298) (0.0202) 2 K c = 0.259

2 H 2 S(g)  2 H 2 (g) + S 2 (g) When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of H 2 S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72×10 –2 mol of S 2 (g) is present at equilibrium. (d)Calculate the partial pressure of S 2 (g) in the container at equilibrium at 483 K. PV = nRT P(1.25L) = 3.72×10 –2 mol (0.08206)(483K) P = 1.18 atm

2 H 2 S(g)  2 H 2 (g) + S 2 (g) When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of H 2 S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72×10 –2 mol of S 2 (g) is present at equilibrium. (e)For the reaction H 2 (g) + ½ S 2 (g)  H 2 S(g) at 483 K, calculate the value of the equilibrium constant, K c. The coefficients are cut in half and the reaction is reversed. K c ’ = K c ’ = 1.96

LeChatelier's Principle When a stress is applied to a system, the equilibrium will shift in the direction that will relieve the stress.

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 a product will cause equilibrium to shift to the left to form more reactants.

A decrease in concentration of a product will cause equilibrium to shift to the right to form more products. A decrease in the concentration of a reactant will cause equilibrium to shift to the left to make more reactants. A change in concentration of reactant or product will not affect the value of K.

If CO is increased, the forward reaction increases to reestablish equilibrium. Therefore the quantity of H 2 will decrease and the quantity of product will increase. The value for the equilibrium constant (K) is unchanged.

If product is added to this system at equilibrium, the reverse reaction will increase to reestablish the equilibrium. Therefore quantities of both reactants (CO and H 2 ) will increase. The value for the equilibrium constant (K) again remains unchanged.

Continuous removal of product from a reaction forces more of it to be produced, according to LeChatelier's Principle. Metabolic reactions as well as industrial processes make use of this effect to continuously make products in equilibrium reactions.

A + B  C + D Add A or B --------> Remove A or B Example: N 2 (g) + 3H 2 (g)  2NH 3 (g) a. addition of N 2 b. addition of NH 3 c. addition of H 2 d. removal of NH 3

Changes in temperature Changes in temperature may easily be treated as changes in concentration if you think of heat as a product (exothermic rxn) or a reactant (endothermic rxn).

An increase in temperature of an exothermic reaction will cause equilibrium to shift to the left. K will decrease. A decrease in temperature of an exothermic reaction will cause equilibrium to shift to the right. K will increase. An increase in temperature of an endothermic reaction will cause equilibrium to shift to the right. K will increase. A decrease in temperature of an endothermic reaction will cause equilibrium to shift to the left. K will decrease.

Changes in pressure 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 of gas on each side, changes in pressure do not affect equilibrium. Adding an inert gas does not affect equilibrium since the partial pressures of the gases in the reaction are not affected. Changing pressure does not affect the value of the equilibrium constant.

Example: P 4 (s) + 6Cl 2 (g)  4PCl 3 (l) a. increase container volume b. decrease container volume c. add argon gas Example: PCl 3 (g) + Cl 2 (g)  PCl 5 (g) a. decrease container volume b. add helium gas Example: PCl 3 (g) + 3NH 3 (g)  P(NH 2 ) 3 (g) + 3HCl(g) a. increase container volume

In the production of ammonia from nitrogen and hydrogen, raising the temperature favors the reverse reaction, which absorbs heat. Temperature and pressure are carefully produced in the industrial production of ammonia, exploiting LeChatelier's Principle to maximize the amount of product obtained. The production of ammonia is of tremendous importance in feeding the world--since ammonia is used as fertilizer.

In the production of ammonia from nitrogen and hydrogen, raising the pressure favors the forward reaction because 4 moles of gas is converted to 2 moles of gas.

Addition of a catalyst Adding a catalyst does not affect equilibrium.

 Example 6. Consider the reaction 2NO 2 (g)  N 2 (g) + 2O 2 (g) which is exothermic. A vessel contains NO 2 (g), N 2 (g), and O 2 (g) at equilibrium. Predict how each of the following stresses will affect the concentration of O 2 and the value of K. A. NO 2 is added B. N 2 is removed C. The volume is halved

D. He(g) is added E. The temperature is increased F. A catalyst is added

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