# M14Q5: Calculations Involving the Equilibrium Constant

Learning Objectives

- Determine the value of an equilibrium constant from equilibrium concentrations.
- Calculate the concentration of a reactant or product at equilibrium when given the value of the equilibrium constant.

| Key Concepts and Summary | Glossary | End of Section Exercises |

While we have learned to identify which direction a reaction will shift to reach equilibrium, we want to extend that understanding to quantitative calculations. We do so by evaluating the ways that the concentrations of products and reactants change as a reaction approaches equilibrium, keeping in mind the stoichiometric coefficients within the reaction. This algebraic approach to equilibrium calculations will be explored in this section.

# Evaluating Stoichiometric Coefficients

Changes in concentrations or pressures of reactants and products occur as a reaction system approaches equilibrium. In this section, we will see that we can relate these changes to each other using the coefficients in the balanced chemical equation describing the system. We use the decomposition of ammonia as an example.

On heating, ammonia reversibly decomposes into nitrogen and hydrogen gases according to this equation:

_{3}(g) ⇌ N

_{2}(g) + 3 H

_{2}(g)

If a sample of ammonia decomposes in a closed system and the concentration of N_{2} increases by 0.11 *M*, the change in the N_{2} concentration is 0.11 M. We can define the change in concentration by Δ[N_{2}], the final concentration minus the initial concentration of N_{2}. The change is positive because the concentration of N_{2} increases.

The change in the H_{2} concentration, Δ[H_{2}], is also positive—the concentration of H_{2} increases as ammonia decomposes. The chemical equation tells us that the change in the concentration of H_{2} is three times the change in the concentration of N_{2} because for each mole of N_{2} produced, 3 moles of H_{2} are produced.

_{2}] = 3 × Δ[N

_{2}] = 3 × (0.11 M) = 0.33 M

The change in concentration of NH_{3}, Δ[NH_{3}], is twice that of Δ[N_{2}]; the equation indicates that 2 moles of NH_{3} must decompose for each mole of N_{2} formed. However, the change in the NH_{3} concentration is negative because the concentration of ammonia *decreases* as it decomposes.

_{3}] = – 2 × Δ[N

_{2}] = -2 × (0.11 M) = -0.22 M

If we did not know the magnitude of the change in the concentration of N_{2}, we could represent it by the variable, *x*. Δ[N_{2}] = *x*

The coefficients in the Δ terms are identical to those in the balanced equation for the reaction.

_{3}(g) ⇌ N

_{2}(g) + 3 H

_{2}(g)

*x*

*x*3

*x*

Note that all the changes on one side of the arrows are of the same sign and that all the changes on the other side of the arrows are of the opposite sign. The simplest way for us to find the coefficients for the concentration changes in any reaction is to use the coefficients in the balanced chemical equation.

### Example 1

**Determining Relative Changes in Concentration**

Complete the changes in concentrations for each of the following reactions.

(a) C_{2}H_{2}(g) + 2 Br_{2}(g) ⇌ C_{2}H_{2}Br_{4}(g)

*x* − −

(b) I_{2}(aq) + I^{–}(aq) ⇌ I_{3}^{–}(aq)

− − *x*

(c) C_{3}H_{8}(g) + 5 O_{2}(g) ⇌ 3 CO_{2}(g) + 4 H_{2}O(g)

*x * − − −

**Solution**

(a) C_{2}H_{2}(g) + 2 Br_{2}(g) ⇌ C_{2}H_{2}Br_{4}(g)

*x* 2*x* −*x*

(b) I_{2}(aq) + I^{–}(aq) ⇌ I_{3}^{–}(aq)

−*x* −*x* *x*

(c) C_{3}H_{8}(g) + 5 O_{2}(g) ⇌ 3 CO_{2}(g) + 4 H_{2}O(g)

*x * 5*x* −3*x* −4*x*

**Check Your Learning**

Complete the changes in concentrations for each of the following reactions:

(a) 2 SO_{2}(g) + O_{2}(g) ⇌ 2 SO_{3}(g)

− *x* −

(b) C_{4}H_{8}(g) ⇌ 2 C_{2}H_{4}(g)

*−2*

*x*

(c) 4 NH_{3}(g) + 7 H_{2}O(g) ⇌ 4 NO_{2}(g) + 6 H_{2}O(g)

*− − −*

### Answer:

(a) 2*x*, *x*, −2*x;* (b) *x*, −2*x;* (c) 4*x*, 7*x*, −4*x*, −6*x* or −4*x*, −7*x*, 4*x*, 6*x*

# Calculation of an Equilibrium Constant

Any problem that requires us to solve for K* _{c}* must provide enough information to determine the reactant and product concentrations at equilibrium. Armed with the concentrations, we can solve the equation for K

*, as it will be the only unknown.*

_{c}The following example shows how to use the stoichiometry of a reaction and a combination of initial and equilibrium concentrations to determine an equilibrium constant. This technique, commonly called an ICE chart—for **I**nitial, **C**hange, and **E**quilibrium–will be helpful in solving many equilibrium problems. A chart is generated beginning with the equilibrium reaction in question. Underneath the reaction the initial concentrations of the reactants and products are listed—these conditions are usually provided in the problem and we consider no shift toward equilibrium to have happened. The next row of data is the change that occurs as the system shifts toward equilibrium—do not forget to consider the reaction stoichiometry as described in a previous section of this chapter. The last row contains the concentrations once equilibrium has been reached.

### Example 2

**Calculation of an Equilibrium Constant**

Iodine molecules react reversibly with iodide ions to produce triiodide ions.

_{2}(aq) + I

^{–}(aq) ⇌ I

_{3}

^{–}(aq)

If a solution with the concentrations of I_{2} and I^{−} both equal to 1.000 × 10^{−3}*M* before reaction gives an equilibrium concentration of I_{2} of 6.61 × 10^{−4}*M*, what is the equilibrium constant for the reaction?

**Solution**

We will begin this problem by calculating the changes in concentration as the system goes to equilibrium. Then we determine the equilibrium concentrations and, finally, the equilibrium constant. First, we set up a table with the initial concentrations, the changes in concentrations, and the equilibrium concentrations using −*x* as the change in concentration of I_{2}.

I_{2}(aq) + I^{–}(aq) ⇌ I_{3}^{–}(aq) |
|||

I: Initial concentration (M) | 1.000 × 10^{-3} |
1.000 × 10^{-3} |
0 |

C: Change (M) | – x |
– x |
+ x |

E: Equilibrium concentration (M) | 1.000 × 10^{-3} – x |
1.000 × 10^{-3} – x |
x |

Since the equilibrium concentration of I_{2} is given, we can solve for *x*. At equilibrium the concentration of I_{2} is 6.61 × 10^{−4}*M* so that

^{-3}–

*x*= 6.61 × 10

^{-4}

*x*= 3.39 × 10

^{-4}M

Now we can fill in the table with the concentrations at equilibrium.

I_{2}(aq) + I^{–}(aq) ⇌ I_{3}^{–}(aq) |
|||

I: Initial concentration (M) | 1.000 × 10^{-3} |
1.000 × 10^{-3} |
0 |

C: Change (M) | – 3.39 × 10^{-4} |
– 3.39 × 10^{-4} |
+ 3.39 × 10^{-4} |

E: Equilibrium concentration (M) | 6.61 × 10^{-4} |
6.61 × 10^{-4} |
3.39 × 10^{-4} |

We now calculate the value of the equilibrium constant.

_{c}= = 776

**Check Your Learning**

Ethanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers.

_{2}H

_{5}OH + CH

_{3}CO

_{2}H ⇌ CH

_{3}CO

_{2}C

_{2}H

_{5}+ H

_{2}O

When 1 mol each of C_{2}H_{5}OH and CH_{3}CO_{2}H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when 0.33 mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is not a solvent in this reaction. This time, it is a reactant dissolved in an organic solvent and therefore should be included in the equilibrium expression.)

### Answer:

K* _{c}* = 4.1

## Calculation of a Missing Equilibrium Concentration

If we know the equilibrium constant for a reaction and know the concentrations at equilibrium of all reactants and products except one, we can calculate the missing concentration.

### Example 3

**Calculation of a Missing Equilibrium Concentration**

Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the equilibrium constant for the reaction, N_{2}(g) + O_{2}(g) ⇌ 2 NO(g), is 4.1 × 10^{−4}. Find the concentration of NO(*g*) in an equilibrium mixture with air at 1 atm pressure at this temperature. In air, [N_{2}] = 0.036 mol/L and [O_{2}] 0.0089 mol/L.

**Solution**

We are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing equilibrium concentration by rearranging the equation for the equilibrium constant.

_{c}=

^{2}= K

_{c}[N

_{2}][O

_{2}]

^{-4}

Thus [NO] is 3.6 × 10^{−4} mol/L at equilibrium under these conditions.

We can check our answer by substituting all equilibrium concentrations into the expression for the reaction quotient to see whether it is equal to the equilibrium constant.

_{c}=

_{c}= 4.0 × 10

^{-4}= K

_{c}

The answer checks; our calculated value gives the equilibrium constant within the error associated with the significant figures in the problem.

**Check Your Learning**

The equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 × 10^{−2}. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 *M* and 2.09 *M*, respectively.

### Answer:

[NH_{3}] = 1.53 M

## Calculation of Changes in Concentration

If we know the equilibrium constant for a reaction and a set of concentrations of reactants and products that are *not at equilibrium*, we can calculate the changes in concentrations as the system comes to equilibrium, as well as the new concentrations at equilibrium. The typical procedure can be summarized in four steps.

- Determine the direction the reaction proceeds to come to equilibrium.
- Write a balanced chemical equation for the reaction.
- If the direction in which the reaction must proceed to reach equilibrium is not obvious, calculate
*Q*from the initial concentrations and compare to K_{c}to determine the direction of change._{c}

- Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.
- Define the changes in the initial concentrations that are needed for the reaction to reach equilibrium. Generally, we represent the smallest change with the symbol
*x*and express the other changes in terms of the smallest change. - Define missing equilibrium concentrations in terms of the initial concentrations and the changes in concentration determined in (a).

- Define the changes in the initial concentrations that are needed for the reaction to reach equilibrium. Generally, we represent the smallest change with the symbol
- Solve for the change and the equilibrium concentrations.
- Substitute the equilibrium concentrations into the equilibrium constant expression, solve for
*x*, and check any assumptions used to find*x*. - Calculate the equilibrium concentrations.

- Substitute the equilibrium concentrations into the equilibrium constant expression, solve for
- Check the arithmetic.
- Check the calculated equilibrium concentrations by substituting them into the equilibrium expression and determining whether they give the equilibrium constant.

Sometimes a particular step may differ from problem to problem—it may be more complex in some problems and less complex in others. However, every calculation of equilibrium concentrations from a set of initial concentrations will involve these steps. In solving equilibrium problems that involve changes in concentration, sometimes it is convenient to set up an ICE table, as described in the previous section.

### Example 4

**Calculation of Concentration Changes as a Reaction Goes to Equilibrium**

Under certain conditions, the equilibrium constant for the decomposition of PCl_{5}(*g*) into PCl_{3}(*g*) and Cl_{2}(*g*) is 0.0211. What are the equilibrium concentrations of PCl_{5}, PCl_{3}, and Cl_{2} if the initial concentration of PCl_{5} was 1.00 *M*?

**Solution**

Use the stepwise process described earlier.

*Determine the direction the reaction proceeds.*The balanced equation for the decomposition of PCl

_{5}isPCl_{5}(g) ⇌ PCl_{3}(g) + Cl_{2}(g)Because we have no products initially,

*Q*= 0 and the reaction will proceed to the right._{c}*Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.*Let us represent the increase in concentration of PCl

_{3}by the symbol*x*. The other changes may be written in terms of*x*by considering the coefficients in the chemical equation.

PCl_{5}(g) ⇌ PCl_{3}(g) + Cl_{2}(g)

*-x*+*x*+*x*The changes in concentration and the expressions for the equilibrium concentrations are:

PCl _{5}(g) ⇌ PCl_{3}(g) + Cl_{2}(g)I: Initial concentration (M) 1.00 0 0 C: Change (M) – *x*+ *x*+ *x*E: Equilibrium concentration (M) 1.00 – *x**x**x**Solve for the change and the equilibrium concentrations.*Substituting the equilibrium concentrations into the equilibrium constant equation gives

K_{c}= = 0.0211 =This equation contains only one variable,

*x*, the change in concentration. We can write the equation as a quadratic equation and solve for*x*using the quadratic formula.0.0211 =0.0211(1.00 –*x*) =*x*^{2}*x*^{2}+ 0.0211*x*– 0.0211 = 0Appendix B shows us an equation of the form

*ax*^{2}+*bx*+*c*= 0 can be rearranged to solve for*x*:*x*=In this case,

*a*= 1,*b*= 0.0211, and*c*= −0.0211. Substituting the appropriate values for*a*,*b*, and*c*yields:*x*=

*x*=*x*= = 0.135 or*x*= = -0.156Quadratic equations often have two different solutions, one that is physically possible and one that is physically impossible (an extraneous root). In this case, the second solution (−0.156) is physically impossible because we know the change must be a positive number (otherwise we would end up with negative values for concentrations of the products). Thus,

*x*= 0.135*M*.The equilibrium concentrations are

[PCl_{5}] = 1.00 – 0.135 = 0.87 M[PCl_{3}] =*x*= 0.135 M[Cl_{2}] =*x*= 0.135 M*Check the arithmetic.*Substitution into the expression for K

(to check the calculation) gives_{c}K_{c}= = 0.021The equilibrium constant calculated from the equilibrium concentrations is equal to the value of K

given in the problem (when rounded to the proper number of significant figures). Thus, the calculated equilibrium concentrations check._{c}

**Check Your Learning**

Acetic acid, CH_{3}CO_{2}H, reacts with ethanol, C_{2}H_{5}OH, to form water and ethyl acetate, CH_{3}CO_{2}C_{2}H_{5}.

_{3}CO

_{2}H + C

_{2}H

_{5}OH ⇌ CH

_{3}CO

_{2}C

_{2}H

_{5}+ H

_{2}O

The equilibrium constant for this reaction with dioxane as a solvent is 4.0 (Note: Water is not a solvent in this reaction. This time it is a reactant dissolved in an organic solvent and therefore should be included in the equilibrium expression). What are the equilibrium concentrations when a mixture that is 0.15 *M* in CH_{3}CO_{2}H, 0.15 *M* in C_{2}H_{5}OH, 0.40 *M* in CH_{3}CO_{2}C_{2}H_{5}, and 0.40 *M* in H_{2}O are mixed in enough dioxane to make 1.0 L of solution?

### Answer:

[CH_{3}CO_{2}H] = 0.36 *M*, [C_{2}H_{5}OH] = 0.36 *M*, [CH_{3}CO_{2}C_{2}H_{5}] = 0.17 *M*, [H_{2}O] = 0.17 *M*

**Check Your Learning**

A 1.00-L flask is filled with 1.00 moles of H_{2} and 2.00 moles of I_{2}. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H_{2}, I_{2}, and HI in moles/L?

_{2}(g) + I

_{2}(g) ⇌ 2 HI(g)

### Answer:

[H_{2}] = 0.06 *M*, [I_{2}] = 1.06 *M*, [HI] = 1.88 *M*

Sometimes it is possible to use chemical insight to find solutions to equilibrium problems without actually solving a quadratic (or more complicated) equation. First, however, it is useful to verify that equilibrium can be obtained starting from two extremes: all (or mostly) reactants and all (or mostly) products.

Consider the ionization of 0.150 M HA, a weak acid.

_{2}O(l) ⇌ H

_{3}O

^{+}(aq) + A

^{–}(aq) K

_{c}= 6.80 × 10

^{-4}

The most obvious way to determine the equilibrium concentrations would be to start with only reactants. This could be called the “all reactant” starting point. Using *x* for the amount of acid ionized at equilibrium, this is the ICE table and solution.

HA(aq) + H_{2}O(l) ⇌ H_{3}O^{+}(aq) + A^{–}(aq) |
||||

I: Initial concentration (M) | 0.150 | – | 0 | 0 |

C: Change (M) | -x | – | +x | +x |

E: Equilibrium Concentration (M) | 0.150 – x | – | x | x |

Setting up and solving the quadratic equation gives

_{c}= = a

*x*

^{2}+ (6.80 × 10

^{-4}

*x*) – 1.02 × 10

^{-4}= 0

*x*=

*x*= 0.00977 M or -0.0104 M

Using the positive (physical) root, the equilibrium concentrations are

*x*= 0.140 M

_{3}O

^{+}] = [A

^{–}] =

*x*= 0.00977 M

Recall that a small K* _{c}* means that very little of the reactants form products and a large K

*means that most of the reactants form products. If the system can be arranged so it starts “close” to equilibrium, then if the change (*

_{c}*x*) is small compared to any initial concentrations, it can be neglected. Small is usually defined as resulting in an error of less than 5%. The following two examples demonstrate this.

### Example 5

**Approximate Solution Starting Close to Equilibrium**

What are the concentrations at equilibrium of a 0.15 *M* solution of HCN?

_{2}O(l) ⇌ H

_{3}O

^{+}(aq) + CN

^{–}(aq) K

_{c}= 4.9 × 10

^{-10}

**Solution**

Using “*x*” to represent the concentration of each product at equilibrium gives this ICE table.

HCN(aq) + H_{2}O(l) ⇌ H_{3}O^{+}(aq) + CN^{–}(aq) |
||||

I: Initial concentration (M) | 0.15 | – | 0 | 0 |

C: Change (M) | -x | – | +x | +x |

E: Equilibrium Concentration (M) | 0.15 – x | – | x | x |

The exact solution may be obtained using the quadratic formula with

_{c}=

solving

*x*

^{2}+ 4.9 × 10

^{-10}

*x*– 7.35 × 10

^{-11}= 0

*x*= 8.6 × 10

^{-6 }M

Thus [H_{3}O^{+}] = [CN^{–}] = *x* = 8.6 × 10^{–6 }*M* and [HCN] = 0.15 – *x* = 0.15 *M*.

In this case, chemical intuition can provide a simpler solution. From the equilibrium constant and the initial conditions, *x* must be small compared to 0.15 *M*. More formally, if *x* << 0.15, then 0.15 – *x* ≈ 0.15. If this assumption is true, then it simplifies obtaining *x*

_{c}= ≈

^{-10}=

*x*

^{2}= (0.15)(4.9 × 10

^{-10}) = 7.4 × 10

^{-11}

*x*= = 8.6 × 10

^{–6 }

*M*

In this example, solving the exact (quadratic) equation and using approximations gave the same result to two significant figures. While most of the time the approximation is a bit different from the exact solution, as long as the error is less than 5%, the approximate solution is considered valid. In this problem, the 5% applies to IF (0.15 – *x*) ≈ 0.15 *M*, so if

is less than 5%, as it is in this case, the assumption is valid. The approximate solution is thus a valid solution.

**Check Your Learning**

What are the equilibrium concentrations in a 0.25 *M* NH_{3} solution?

_{3}(aq) + H

_{2}O(l) ⇌ NH

_{4}

^{+}(aq) + OH

^{–}(aq) K

_{c}= 1.8 × 10

^{-5}

Assume that *x* is much less than 0.25 M and calculate the error in your assumption.

### Answer:

[OH^{–}] = [NH_{4}^{+}] = 0.0021 M; [NH_{3}] = 0.25 *M*, error = 0.84%

The second example requires that the original information be processed a bit, but it still can be solved using a small *x* approximation.

### Example 6

**Approximate Solution After Shifting Starting Concentration**

Copper(II) ions form a complex ion in the presence of ammonia

^{2+}(aq) + 4 NH

_{3}(aq) ⇌ Cu(NH

_{3})

_{4}

^{2+}(aq) K

_{c}= 5.0 × 10

^{13}=

If 0.010 mol Cu^{2+} is added to 1.00 L of a solution that is 1.00 *M* NH_{3} what are the concentrations when the system comes to equilibrium?

**Solution**

The initial concentration of copper(II) is 0.010 *M*. The equilibrium constant is very large so it would be better to start with as much product as possible because “all products” is much closer to equilibrium than “all reactants.” Note that Cu^{2+} is the limiting reactant; if all 0.010 *M* of it reacts to form product the concentrations would be

^{2+}] = 0.010 – 0.010 = 0 M

_{3})

_{4}

^{2+}] = 0.010 M

_{3}] = 1.00 – 4 × 0.010 = 0.96 M

Using these “shifted” values as initial concentrations with *x* as the free copper(II) ion concentration at equilibrium gives this ICE table.

Cu^{2+}(aq) + 4 NH_{3}(aq) ⇌ Cu(NH_{3})_{4}^{2+}(aq) |
|||

I: Initial concentration (M) | 0 | 0.96 | 0.010 |

C: Change (M) | + x |
+ 4x |
– x |

E: Equilibrium concentration (M) | + x |
0.96 + 4x |
0.010 – x |

Since we are starting close to equilibrium, *x* should be small so that

*x*≈ 0.96 M

*x*≈ 0.010 M

_{c}= ≈ = 5.0 × 10

^{13}

*x*= = 2.4 × 10

^{-16 }M

Select the smallest concentration for the 5% rule.

^{-12}%

This is much less than 5%, so the assumptions are valid. The concentrations at equilibrium are

^{2+}] =

*x*= 2.4 × 10

^{-16 }M

_{3}] = 0.96 – 4

*x*= 0.96 M

_{3})

_{4}

^{2+}] = 0.010 – x = 0.010 M

By starting with the maximum amount of product, this system was near equilibrium and the change (*x*) was very small. With only a small change required to get to equilibrium, the equation for *x* was greatly simplified and gave a valid result well within the 5% error maximum.

**Check Your Learning**

What are the equilibrium concentrations when 0.25 mol Ni^{2+} is added to 1.00 L of 2.00 *M* NH_{3} solution?

^{2+}(aq) + 6 NH

_{3}(aq) ⇌ Ni(NH

_{3})

_{6}

^{2+}(aq) K

_{c}= 5.5 × 10

^{8}

With such a large equilibrium constant, first form as much product as possible, then assume that only a small amount (*x*) of the product shifts left. Calculate the error in your assumption.

### Answer:

[Ni(NH_{3})_{6}^{2+}] = 0.25 M, [NH_{3}] = 0.50 *M*, [Ni^{2+}] = 2.9 × 10^{–8}*M*, error = 1.2 × 10^{–5}%

## Key Concepts and Summary

In this section, we dive into the types of calculations that are common in systems at equilibrium. Most of these types of questions require you to first write the chemical reaction, then set up an ICE chart, then setting up at K_{eq} expression using the values in “E” of the ICE chart. In these types of problems, there are three primary variables – the value of K_{eq}, the initial concentrations, and the equilibrium concentrations. Most often, two of these variables are measured and the third is unknown and needs to be calculated. If a system is not initially at equilibrium, we can calculate Q and when compared to K_{eq}, we know which way the reaction will shift in order to achieve equilibrium. This shift is reflected in “C” of the ICE chart – the side the equilibrium shifts to contains the “+ x”.

## Glossary

**ICE chart**

Initial, Change, Equilibrium – a chart helpful in solving many equilibrium problems

### Chemistry End of Section Exercises

- A 0.72-mol sample of PCl
_{5}is put into a 1.00 L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl_{3}(g) and 0.40 mol of Cl_{2}(g). Calculate the value of the equilibrium constant for the decomposition of PCl_{5}to PCl_{3}and Cl_{2}at this temperature. - A sample of ammonium chloride was heated in a closed container:

NH_{4}Cl(s) ⇌ NH_{3}(g) + HCl(g)

At equilibrium, the pressure of NH_{3}(*g*) was found to be 1.75 atm. What is the value of the equilibrium constant,*K*, for the decomposition at this temperature?_{P} - What is the pressure of BrCl in an equilibrium mixture of Cl
_{2}, Br_{2}, and BrCl if the equilibrium pressure of Cl_{2}is 0.115 atm and the equilibrium pressure of Br_{2}is 0.450 atm?

Br_{2}(g) + Cl_{2}(g) ⇌ 2 BrCl(g) K_{p}= 4.7 × 10^{-2} - Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H
_{2}and 1.25 mol of I_{2}in a 5.00 L flask at 448 °C.

H_{2}(g) + I_{2}(g) ⇌ 2 HI(g) K_{c}= 50.2 at 448°C - A student solved the following problem:

*The problem was: For the following reaction at 600 °C: For the following reaction at 600 °C:*

2 SO_{2}(g) + O_{2}(g) ⇌ 2 SO_{3}(g) K_{c}= 4.32

and found the equilibrium concentrations to be [SO_{2}] = 0.590 M, [O_{2}] = 0.0450 M, and [SO_{3}] = 0.260 M. How could this student check the work without reworking the problem? - What are the equilibrium concentrations after a mixture in a sealed container, initially containing [H
_{2}O] = 1.00 M and [Cl_{2}O] = 1.00 M, comes to equilibrium at 25°C?

H_{2}O(g) + Cl_{2}O(g) ⇌ 2 HOCl(g) K_{c}= 0.0900 - What are the concentrations of PCl
_{5}, PCl_{3}, and Cl_{2}in an equilibrium mixture produced by the decomposition of a sample of pure PCl_{5}with [PCl_{5}] = 2.00 M? Assume a constant temperature.

PCl_{5}(g) ⇌ PCl_{3}(g) + Cl_{2}(g) K_{c}= 0.0211 - What is the minimum mass of CaCO
_{3}required to establish equilibrium at a certain temperature in a 6.50 L container if the equilibrium constant (*K*) is 0.050 for the decomposition reaction of CaCO_{c}_{3}at that temperature?

CaCO_{3}(s) ⇌ CaO(s) + CO_{2}(g) K_{c}= 0.050 - In a 3.0 L vessel, the following equilibrium partial pressures are measured: N
_{2}, 190. torr; H_{2}, 317 torr; NH_{3}, 1.00 × 10^{3}torr.

N_{2}(g) + 3 H_{2}(g) ⇌ 2 NH_{3}(g)- How will the partial pressures of H
_{2}, N_{2}, and NH_{3}change if H_{2}is removed from the system? Will they increase, decrease, or remain the same? - Hydrogen is removed from the vessel until the partial pressure of hydrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.

- How will the partial pressures of H
- The equilibrium constant (
*K*) for this reaction is 5.0 at a given temperature:_{c}

CO(g) + H_{2}O(g) ⇌ CO_{2}(g) + H_{2}(g)- On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H
_{2}in a liter. How many moles of CO_{2}were there in the equilibrium mixture? - Maintaining the same temperature, additional H
_{2}was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of CO, 0.30 mol of water vapor, and 1.2 mol of H_{2}in a liter. How many moles of CO_{2}were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.

- On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of CO, 0.30 mol of water vapor, and 0.90 mol of H

### Answers to Chemistry End of Section Exercises

- K
_{c}= 0.50 - K
_{p}= 3.06 - P
_{BrCl}= 0.0493 atm - 8.86 moles
- Calculate
*Q*based on the calculated concentrations and see if it is equal to*K*. Because_{c}*Q*does equal 4.32, the system must be at equilibrium. - [H
_{2}O] = 0.870 M, [Cl_{2}O] = 0.870 M, [HOCl] = 0.260 M - [PCl
_{5}] = 1.81, [PCl_{3}] = 0.195 M, [Cl_{2}] = 0.195 M - 32.5 grams
- (a) Removing H
_{2}gas from the system will disturb the equilibrium, the resulting Q_{P}value will will larger than K_{p}. This will cause the equilibrium to favor the reverse reaction until equilibrium is reestablished. After equilibrium is reestablished, the pressure of N_{2}will be higher and the pressure of the NH_{3}will be lower than prior to removing the hydrogen. The hydrogen pressure will go down quickly and then increase, but not to a pressure as high as before the removal so it will be lower. (b) K_{p}= 1.65 × 10^{-4}; P_{N2}= 168 torr, P_{NH3}= 955 torr - (a) 0.33 moles; (b) CO
_{2}= 0.50 moles*;*Added H_{2}forms some water to compensate for the removal of water vapor and as a result of a shift to the left after H_{2}is added.

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