Student Worksheet for Equilibrium€¦ · Advanced Chemistry Equilibrium © 2017 Supercharged...

Advanced Chemistry Equilibrium

© 2017 Supercharged Science www.ScienceLearningSpace.com 1

Student Worksheet for Equilibrium

Attempt to work the following practice problems after working through the sample problems in

the videos. Answers are given on the last page(s).

Relevant Equations

# of moles = 𝐺𝑟𝑎𝑚𝑠 𝐺𝑖𝑣𝑒𝑛

𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑊𝑒𝑖𝑔ℎ𝑡

Keq = 𝐶𝑐𝐷𝑑

𝐴𝑎𝐵𝑏

Quadratic Equation: Ax2 + Bx + C = 0

Where, denominators represent the reactants and numerators represent the products, exponents are

the coefficients from the ICE table, and Keq is the equilibrium constant.



1. State Le Chatelier’s Principle in general terms, and the three variables that can shift equilibrium.

2. Explain why a catalyst does not affect equilibrium point.

3. For each of the following conditions, identify how equilibrium will shift.

4 HNO3 + O2 2 H2O + 4 NO3

Action Effect on HNO3 Effect on O2 Effect on NO3 Products or Reactants Favored?

Addition of HNO3

Addition of O2

Removal of Water

Increase Pressure

Increase Temperature

4. A 0.75 mol sample of dinitrogen tetroxide is placed in a 1.50 L vessel and heated to 283K. At equilibrium it is found that 24.5 % of the dinitrogen tetroxide has decomposed to nitrogen dioxide. Calculate the Keq for this reaction.

N2O4 2 NO2



5. The following reaction has Keq value of 76.0 at 300°C:

HSO4- + CH3OH H2SO4 + CH3O-

If a mixture that contains 1 mol of each is prepared, what is the concentration of each reactant at

equilibrium at 300°C?

6. If the equilibrium constant for the following reaction is 0.97, what amount of CaCl2 will remain at

equilibrium if 1 mole of Calcium Chloride is dissolved in 2 moles of water?

CaCl2 + 2 H2O 2 HCl + Ca(OH)2

7. During the Iodine-Starch reaction, it was mentioned that increasing the concentration of KI would

cause a faster reaction while increasing the concentration of NaS2O3 would cause a slower color change.

Using the following reaction equations, explain this observation.

Step 1: I3- + 2 S2O3

2- 3 I- + S4O62-

Step 2: 2 I3- + Starch Starch-I5- + I-



8. At 350°C, the equilibrium constant for the reaction between hydrogen and nitrogen gases to make

ammonia gas is 43.5. A sample of ammonia was added to a 1.50 L vessel and it was found that at equilibrium 0.218 mol of H2 gas was present. How many moles of NH3 were originally placed into the flask?

N2 + 3 H2 2 NH3

9. At 275°C the equilibrium constant for the following reaction was found to be 0.25. What is the equilibrium concentration of H2CO3 at this temperature?

NaHCO3 + HCl NaCl + H2CO3

10. Will the reaction in #9 ever reach equilibrium if 50g of NaHCO3 are added? Explain your answer.



11. If the equilibrium constant of the following equation is 23.7 at 100˚C, how many moles of SO2 need

to be added to 3 moles of NO2 to reach equilibrium of NO?

SO2 + NO2 SO3 + NO

12. Phosphorus pentachloride decomposes into phosphorus trichloride and chlorine gas according to the following equation. If 21.2 % of a 1.5 mol/L concentration of phosphorus pentachloride decomposes, find the value of Keq.

PCl5 PCl3 + Cl2

13. Find the equilibrium constant if 4.50 mol of A2O3 and 1.37 mol of O2 are initially present in a 1.0 L volume according to the following equation. Assume that Al2O3 has a 25% disappearance at equilibrium.

4 AO2 2 A2O3 + O2



14. The Keq value for the formation of carbon dioxide gas from methane and oxygen gas is 2.4. If the equilibrium concentration of oxygen was 4.50 M, determine the number of moles of carbon dioxide originally present in a 1.0 L vessel.

2 CH4 + 10 O2 8 CO2 + 4 H2O

15. Based on your answers and information in #14, will equilibrium be reached if 35 grams of CO2 is dissolved in 100mL of water?



1. State Le Chatelier’s Principle in general terms, and the three variables that can shift equilibrium.

The principle describes how a system recovers when it’s previous equilibrium state as been disturbed.

The equilibrium of a system can be shifted by changes in pressure, temperature, and/or concentrations.

2. Explain why a catalyst does not affect equilibrium point.

A catalyst speeds up a chemical reaction by having a lower activation energy. Equilibrium is the point in

a system that the concentration of reactants equals the concentration of products. Hence, the catalysts

may cause equilibrium to be reached faster, but it does not change the concentration balance of the

system.

3. For each of the following conditions, identify how equilibrium will shift.

4 HNO3 + O2 2 H2O + 4 NO3

Action Effect on HNO3 Effect on O2 Effect on NO3 Products or Reactants Favored?

Addition of HNO3 N/A Decrease Increase Products

Addition of O2 Decrease N/A Increase Products

Removal of Water Decrease Decrease Increase Products

Removal of O2 Increase N/A Decrease Reactants

Increase Pressure Decrease Decrease Increase Products

Increase Temperature Increase Increase Decrease Reactants

4. A 0.75 mol sample of dinitrogen tetroxide is placed in a 1.50 L vessel and heated to 283K. At equilibrium it is found that 24.5 % of the dinitrogen tetroxide has decomposed to nitrogen dioxide. Calculate the Keq for this reaction.

N2O4 2 NO2

N2O4 NO2

I 0.75 0

C -X +2X

E 0.75-X 2X

The problem tells us that 24.5% of the N2O4 is gone, so it is the value of X.

X= 0.75*0.245 = 0.18 moles

The amount of N2O4 remaining = 0.57 moles

The amount of NO2 present = 0.36 moles

Keq = 𝐵𝑏

𝐴𝑎 = 0.362

0.57 = 0.23



5. The following reaction has Keq value of 76.0 at 300°C:

HSO4- + CH3OH H2SO4 + CH3O-

If a mixture that contains 1 mol of each is prepared, what is the concentration of each reactant at

equilibrium at 300°C?

HSO4- CH3OH H2SO4 CH3O-

I 1 1 0 0

C -X - X + X + X

E 1-X 1-X +X +X


𝐴𝑎𝐵𝑏

76= 𝑥2

(1−𝑋)(1−𝑋)

Because there are the same number of moles, take the square root of both sides. Note that this can not

be done when all values are not equal.

8.72 = 𝑋

1−𝑋 Multiply both sides by the denominator.

8.72-8.72X = X

8.72 = 9.72X

X = 0.90 moles

At equilibrium, the concentrations are:

HSO4- : 0.10 moles

CH3OH: 0.10 moles

H2SO4: 0.9 moles

CH3O-: 0.9 moles

6. If the equilibrium constant for the following reaction is 0.97, what amount of CaCl2 will remain at

equilibrium if 1 mole of Calcium Chloride is dissolved in 2 moles of water?

CaCl2 + 2 H2O 2 HCl + Ca(OH)2

CaCl2 H2O HCl Ca(OH)2

I 1 2 0 0

C - X -X + 2X + X

E 1-X 2-X 2X 2X




𝐴𝑎𝐵𝑏

0.97 = 2𝑋∗𝑋

(1−𝑋)(2−𝑋) Multiply both sides by the denominator.

0.97(1-X)(2-X) = 2X2 FOIL, Distribute and Combine like terms.

0.97X2 – 1.94X + 1.94 = 2X2 Set everything equal to 0, and use the Quadratic Eq. to solve for X.

X= 0.72, -2. Only use the positive number.

Plugging the value into the E line of the ICE table, we find that:

0.38 moles of CaCl2 are present at equilibrium.

7. During the Iodine-Starch reaction, it was mentioned that increasing the concentration of KI would

cause a faster reaction while increasing the concentration of NaS2O3 would cause a slower color change.

Using the following reaction equations, explain this observation.

Step 1: I3- + 2 S2O3

2- 3 I- + S4O62-

Step 2: 2 I3- + Starch Starch-I5- + I-

The first thing that you should notice from Step two is that the formation of S4O62- is required to form

the required ion of iodine to interact with the starch. This I ion does not finish forming until all of the

thiolsulfate has reacted. If you have a constant concentration of thiolsulfate, but more I, the thiolsulfate

takes less time to “find” the I3-. On the contrary, if you increase the thiolsulfate with a constant

concentration of I, it will take longer for all of the thiolsulfate to react, which prolongs the formation of

I3-.

Note the I reactants in Step 1 and 2. They are similar in formula, but very different in structure.

8. At 350°C, the equilibrium constant for the reaction between hydrogen and nitrogen gases to make

ammonia gas is 43.5. A sample of ammonia was added to a 1.50 L vessel and it was found that at equilibrium 0.218 mol of H2 gas was present. How many moles of NH3 were originally placed into the flask?

N2 + 3 H2 2 NH3

The very first thing you should notice about this problem is that we are using Le Chatelier’s Principle to drive the reaction in favor of reactants. This is because only the product is put into the vessel, so reactants are formed to establish equilibrium (reverse of normal, but practical).

N2 H2 NH3

I 0 0 2Y

C + X + 3X -X

E X 3X 2Y-X

If at equilibrium, the [H2] was 0.218 moles, the [N2] was 0.218 𝑚𝑜𝑙𝑒𝑠

3= 7.3 ∗ 102 𝑚𝑜𝑙𝑒𝑠 because of the

3:1 ratio between H2 and N2. Therefore, X= 0.073.



Keq = 𝐶𝑐

𝐴𝑎𝐵𝑏

43.5 = (2𝑌−0.073)2

0.073∗(3∗0.073)3

43.5 = 4𝑌2−0.0053

7.67∗10−4 Solve for Y.

0.03= 4Y2 - 0.0053

Y= 0.0053 moles of NH3 were originally added

9. At 275°C the equilibrium constant for the following reaction was found to be 0.25. What is the equilibrium concentration of H2CO3 at this temperature?

NaHCO3 + HCl NaCl + H2CO3

In this case, everything is in a 1:1 ratio, so there is no need to do an ICE table. The Keq is the [H2CO3]. You may do an ICE table if you’d like, or refer to #5 that had the same conditions.

[H2CO3] = 0.25 moles

10. Will the reaction in #9 ever reach equilibrium if 50g of NaHCO3 are added? Explain your answer.

Simply convert grams to moles.

# of moles= 𝐺𝑟𝑎𝑚𝑠 𝑔𝑖𝑣𝑒𝑛


= 50 𝑔

84𝑔/𝑚𝑜𝑙

= 0.60 moles of NaHCO3

Because there is a 1:1 ratio, this tells you that at this temperature, it is possible to produce 0.60 moles of

H2CO3.

The [H2CO3] is 0.25 moles, so equilibrium will be reached.

***If the number of moles were less than the concentration at equilibrium, equilibrium would not be

reached. *****

11. If the equilibrium constant of the following equation is 23.7 at 100˚C, how many moles of SO2 need

to be added to 3 moles of NO2 to reach equilibrium of NO?

SO2 + NO2 SO3 + NO

Again, acknowledge the 1:1, so there is no need to do an ICE table.

Keq = 𝐶𝑐

𝐴𝑎𝐵𝑏

23.7 = 2

𝐴3

71.1A= 2



[A] = 2.8*10-2 moles of NO2

12. Phosphorus pentachloride decomposes into phosphorus trichloride and chlorine gas according to

the following equation. If 21.2 % of a 1.5 mol/L concentration of phosphorus pentachloride

decomposes, find the value of Keq.

PCl5 PCl3 + Cl2

PCl5 PCl3 Cl2

I 1 0 0

C - X + X + X

E 1-X X X

Moles of PCl5 that are removed= 0.212*1.5= 0.32 moles PCl5. This is also the value of X.

Keq = 𝐵𝑏𝐶𝑐

𝐴𝑎

= 0.322

1−0.32

= 0.15

13. Find the equilibrium constant if 4.50 mol of Al2O3 and 1.37 mol of O2 are initially present in a 1.0 L volume according to the following equation. Assume that Al2O3 has a 25% disappearance at equilibrium.

4 AlO2 2 Al2O3 + O2

This problem is colored coordinated so that you can follow.

AlO2 Al2O3 O2

I 0 4.5 1.37

C + 4X - 2X - X

E 2.28 4.5- 1.13 = 3.37 1.37- 0.57= 0.80

4.5 mole Al2O3 * 0.25 = 1.13 moles disappeared

Moles of O2 that disappeared = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐴𝑙2𝑂3

2 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑜𝑓 2: 1 𝑟𝑎𝑡𝑖𝑜 =

1.13

2= 0.57 𝑚𝑜𝑙𝑒𝑠 𝑂2

The calculation for O2 reveals X, which is 0.57 moles; hence, the value of AlO2 produce is 4*0.57 moles

Keq = 𝐵𝑏𝐶𝑐

𝐴𝑎

= (3.37∗0.80)

2.28

= 1.18



14. The Keq value for the formation of carbon dioxide gas from methane and oxygen gas is 2.4. If the equilibrium concentration of oxygen was 4.50 M, determine the number of moles of carbon dioxide originally present in a 1.0 L vessel.

CH4 + 5 O2 4 CO2 + 2 H2O

CH4 O2 CO2

I 0 0 Y

C + X + 5X -8X

E X 5X Y-4X

If at equilibrium, the [O2] was 0.450 moles, the [CH4] was 0.450 𝑚𝑜𝑙𝑒𝑠

5= 9.0 ∗ 102 𝑚𝑜𝑙𝑒𝑠 because of the

5:1 ratio between O2 and CH4. Therefore, X= 0.090.

Keq = 𝐶𝑐

𝐴𝑎𝐵𝑏

43.5 = (𝑌−4𝑋)2

0.073∗(5∗0.073)5

43.5 = 𝑌2−0.032

5.0∗10−9 Solve for Y.

2.1 *10 -7 = Y- 0.032

Y= 0.032 moles of CO2 were originally added

15. Based on your answers and information in #14, will equilibrium be reached if 35 grams of CO2 is dissolved in 100mL of water?

# of moles = 𝐺𝑟𝑎𝑚𝑠 𝐺𝑖𝑣𝑒𝑛


= 46

35 = 1.31 moles CO2

Equilibrium will be reached.

Student Worksheet for Equilibrium€¦ · Advanced Chemistry Equilibrium © 2017 Supercharged...

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