# 7B: Kinetics to Equilibrium (Worksheet)

- Page ID
- 81627

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Name: ______________________________

Section: _____________________________

Student ID#:__________________________

Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.

Most chemical reactions are reversible. This means that once products are formed, they can react to reform the reactants. If we allow a reaction to run long enough, it may reach a state where the rate of the forward reaction (forming products) is equal to the rate of the reverse reaction (reforming reactants). When this occurs, the reaction is said to be in *equilibrium*. At equilibrium, reactant and product amounts do not change over time, and they maintain a fixed ratio, which can be expressed as an *equilibrium constant*. The concept of the equilibrium constant is one of the most powerful in chemistry, allowing us to calculate amounts of products and remaining reactants in real reaction mixtures.

## Learning Objectives

- Understand the meaning of dynamic equilibrium
- Understand how the differential rate expressions for the forward and reverse processes lead to the definition of the equilibrium constant, K, for the reaction
- Understand how the equilibrium constant, \(K\), is defined from a balanced chemical equation for a reaction
- Understand how to assess changes in reactant and product amounts and to set up calculations of amounts present at equilibrium

## Success Criteria

- Be able to write the Kc expression for any balanced reaction equation
- Be able to calculate the value of Kc from equilibrium concentration data and stoichiometry
- Be able to use the stoichiometry of a balanced equation to write algebraic expressions for concentrations of reactants and products at equilibrium
- Be able to calculate equilibrium concentrations, given the value of Kc and initial amounts of reactants and/or products.

## Dynamic Equilibrium and the Equilibrium Constant

As a chemical reaction runs to convert reactants into products, the build-up of products increases the likelihood that they will react with each other to reform the reactants. In other words, most real reactions do not run "to completion," meaning total conversion of reactants into products. In the early stages of a chemical reaction, the product concentrations are small, so the rate of the reverse reaction is slow. But as product amounts increase, the rate of the reverse reaction increases. At the same time, as reactants are consumed, the rate of the forward reaction is slowing. Eventually, there comes a point at which the rate of the forward reaction exactly equals the rate of the reverse reaction. This is the point of *dynamic equilibrium*. The equilibrium state is dynamic, because both the forward and reverse processes are still taking place. Nonetheless, at equilibrium the amounts of reactants and products in the reaction mixture do not change over time. At equilibrium, as soon as products are formed they react by the reverse reaction to reform reactants, and vice versa. Because the amounts of reactants and products do not change over time, there exists a fixed ratio between their amounts. We can define this ratio, called the equilibrium constant, \(K\), in such a way that there is a connection between its form and the stoichiometry of the balanced equation. If the amounts of all species are stated in concentrations of mol/L, we call this constant \(K_c\). The following steps will take you though the definition of an equilibrium constant for a simple reaction and show you the relationship between its form and the stoichiometry of the balanced equation.

### Q1

Consider the following reversible reaction, which is believed to proceed by a one-step mechanism in each direction:

\[ \ce{2NO2 <=>[k_f][k_r] N2O4} \nonumber \]

- Write the rate expression for each direction.
- At equilibrium, the net rate of the reaction is zero, because the rate of the forward reaction is exactly equal to that of the reverse reaction. Set your two rate expressions equal to each other, and solve the equality for \(\frac{k_f}{k_r}\). This ratio of constants is itself a constant, which we call the
*equilibrium constant*, \(K_c\).

### Defining the Form of *K*_{c} from a Balanced Equation

_{c}

Note that the expression you have derived for \(K_c\) is a ratio between the concentration of product, [N_{2}O_{4}], and the concentration of reactant, [NO_{2}], at equilibrium:

\[ K_c = \dfrac{[N_2O_4]}{[NO_2]^2} \nonumber \]

Moreover (and most importantly), each concentration term is raised to a power that is its stoichiometric coefficient in the balanced reaction equation (here, an unwritten 1 for [N_{2}O_{4}] and a written 2 for [NO_{2}]). We will not prove it here, but for a general reaction of the form

\[a A + b B \rightleftharpoons c C + d D \nonumber \]

where the lower case letters represent the stoichiometric coefficients, at equilibrium the concentrations of all species will have specific values that define a constant of the form

\[ K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \nonumber \]

regardless of the number of steps in the mechanism. Thus, we can write down the form of \(K_c\) by simply noting the stoichiometry of the balanced equation. In doing this, the following conventions are observed:

- For \(K_c\), all concentrations are the values
*at equilibrium*in mol/L. - \(K_c\) is defined as the ratio of equilibrium concentrations of products over reactants, each raised to powers equal to their stoichiometric coefficients in the balanced reaction equation.
- All solids and neat liquids (not solutions) behave as if they have unit concentration and do not appear in the expression for \(K_c\).
- For solutions, solutes appear in the expression for \(K_c\), but solvents are usually omitted, unless the solvent’s concentration could change significantly through active participation in the reaction.
- Non-reactive species, such as spectator ions, are omitted from \(K_c\).
*K*is defined for the reaction proceeding in the usual left-to-right manner, as written. The value of \(K_c\) for the reverse reaction is the inverse of that for the forward reaction. For example_{c}

### Q2

Write the expressions for the equilibrium constant, *K _{c}* for the following reactions.

- \(\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\)
- \(\ce{Cu(NO3)2(aq) + Zn(s) <=> Cu(s) + Zn(NO3)2(aq)}\)
*[Hint: Write the net ionic equation first.]* - \(\ce{CaO(s) + CO2(g) <=> CaCO3(s)}\)
- \(\ce{CaCO3(s) <=> CaO(s) + CO2(g)}\) How does this compare to the expression you wrote for the reverse reaction in part c?
- \(\ce{HF(aq) + H2O(l) _ H3O+(aq) + F^{–}(aq)}\)
*[Hint: \(\ce{H2O(l)}\) is a solvent.]*

## Relationship between \(K\) and the form of the balanced equation

As we see, the form of the equilibrium expression depends upon the stoichiometric coefficients used to balance the reaction equation. In most cases, we tend to balance equations in terms of lowest whole number coefficients. But occasionally it is more sensible to balance the equation with either higher whole-number coefficients or even fractional coefficients. (Recall that we saw this frequently in our discussions of thermodynamics.) The way in which we choose to balance the equation determines the form of the equilibrium constant and therefore its numerical value according to how it is defined. For example, the equilibrium constant for the following reaction,

\[2 NO(g) + Br_2(g) \rightleftharpoons 2 NOBr(g) \nonumber \]

is \(K_c = 1.3 \times 10^{–2}\) at 1000 K. This value relates to the equilibrium expression defined on the basis of this way of balancing the reaction equation; i.e.,

\[ K_c = \dfrac{[NOBr]^2}{[NO]^2[Br_2]} = 1.3 \times 10^{-2} \nonumber \]

If for some reason we balanced the reaction equation as follows,

\[\ce{4 NO(g) + 2 Br2(g) <=> 4 NOBr(g)} \nonumber \]

the related equilibrium expression would be

\[ K_c = \dfrac{[NOBr]^4}{[NO]^4[Br_2]^2} \nonumber \]

This is the square of our previously defined \(K_c\), so its value is \(K'_c = (K_c)^2 = 1.7 \times 10^{–4}\).

### Q3

For the equilibrium

\[\ce{2 NO(g) + 2 H2(g) <=> N2(g) + 2 H2O(g)} \nonumber \]

with \(K_c = 654\) at 300 K.

- Write the expression for \(K_c\) on the basis of the given balanced equation.
- Suppose the equation were balanced as follows: \[\ce{NO(g) + H2(g) <=> ½ N2(g) + H2O(g)} \nonumber \] Write the expression for \(K_c\) on the basis of this form of the balanced equation and calculate its value.

## Using Equilibria and their Constants to calculate other Equilibrium Constants

We can use equilibrium reaction equations and their associated equilibrium constants in much the same way that we have seen with Hess’s Law, where we used thermochemical equations to find the enthalpy value for some overall equation. The reactions and K values used may constitute a purely hypothetical sum to get an overall value, or they may represent actual equilibria that are occurring simultaneously in the same reaction mixture. For example, suppose we wish to know the value of the equilibrium constant for the reaction at 298 K

\[\ce{N2(g) + O2(g) + Br2(g) <=> 2 NOBr(g)} \nonumber \]

given the following two equilibria and their constants at the same temperature:

Chemical Reaction |
Equilibrium Constant |
---|---|

\(\ce{2 NO(g) <=> N2(g) + O2(g) }\) | \(K_c = 2.1 \times 10^{30}\) |

\(\ce{2 NO(g) + Br2(g) <=> 2 NOBr(g) }\) | \(K_c = 2.0\) |

We can see the target equilibrium as the sum of the reverse of the first given reaction plus the second given reaction as shown; i.e.

Chemical Reaction |
Equilibrium Constant |
---|---|

\(\ce{N2(g) + O2(g) <=> 2 NO(g)}\) | \(K'_c = \dfrac{1}{K_c} = \dfrac{1}{2.1 \times 10^{30}} = 4.76 \times 10^{–31}\) |

\(\ce{2 NO(g) + Br2(g) <=> 2 NOBr(g)}\) | \(K_c = 2.0\) |

\(\ce{N2(g) + O2(g) + Br2(g) <=> 2 NOBr(g)}\) | \(K_c = (2.0)(4.76 \times 10^{–31}) = 9.5 \times 10^{–31}\) |

The \(K_c\) values for each equilibrium in the sum are those appropriate to the ways in which they are written. Note that \(K'_c\) for the first reaction in the sum is the inverse of the given value, \(1/K_c\), because it is being used in the reverse direction. The value of \(K_c\) for the overall equilibrium is the product of the values for the individual steps. We can show this by multiplying the equilibrium expression for the two steps and seeing that by cancelling the result is the expression for the overall equilibrium:

\[ \begin{align} K_c &= \left( \dfrac{ \cancel{[NO]^2}}{[N_2][O_2]}\right) \left( \dfrac{[NOBr]^2}{\cancel{[NO]^2}[Br_2]}\right) = \dfrac{ [NOBr]^2 }{[N_2][O_2][Br_2]} \\ &= \dfrac{2.0}{2.1 \times 10^{30}} = 9.5 \times 10^{-31} \end{align} \nonumber \]

### Q4

Given the following equilibria and their \(K_c\) values at 1200 K:

\[\ce{CO(g) + 3 H2(g) <=> CH4(g) + H2O(g) } \nonumber \]

with \(K_c(1) = 3.9 \)

\[\ce{CS2(g) + 4 H2(g) <=> CH4(g) + 2 H2S(g) } \nonumber \]

with \(K_c(2) = 3.0 \times 10^{–5}\)

Calculate the value of \(K_c\) for the equilibrium

\[\ce{CO(g) + 2 H2S(g) <=> CS2(g) + H2O(g) + H2(g)} \nonumber \]

## The Value of K and its Meaning

Suppose we allow a certain reaction to reach equilibrium at a particular temperature. If we then determine the concentrations of all reactant and product species in the equilibrium mixture, we can substitute these values into the \(K_c\) expression to calculate the numerical value of the equilibrium constant for the reaction at that temperature. If we run the reaction again at the same temperature, but with different starting concentrations, the individual concentrations of reactants and products at equilibrium will be different from those in the first case. But the ratio between product and reactant concentrations, as defined by \(K_c\), will be the same in both cases.

The magnitude of the \(K_c\) value indicates whether the reaction at equilibrium favors products or reactants. The ratio that defines \(K_c\) places product species concentrations in the numerator and reactant species concentrations in the denominator. Therefore, if \(K_c\) for a reaction is greater than 1, at equilibrium products will predominate. The reaction is said to be product favored, or alternatively it is said to lie to the right. Conversely, if \(K_c\) is less than 1, then at equilibrium reactants will predominate. In this case the reaction is said to be reactant favored, or alternatively it is said to lie to the left.

### Q5

Consider the following reaction

\[\ce{H2(g) + I2(g) <=> 2 HI(g)} \nonumber \]

At 425 °C, an equilibrium mixture has the following concentrations

- \([\ce{HI}] = 1.01 \times 10^{-2} mol/L \)
- \([\ce{H2}] = 1.25 \times 10^{-3} mol/L \)
- \([\ce{I2}] = 1.49 \times 10^{-3} mol/L \)

- What is the value of \(K_c\)?
- In general, does an equilibrium mixture of \(\ce{H2(g)}\), \(\ce{I2(g)}\), and \(\ce{HI(g)}\) at 425 °C contain mostly reactants or mostly products?
- In another experiment at 425 °C, an equilibrium mixture is found to contain 0.25 mol/L of \(\ce{H2(g)}\) and 0.50 mol/L of \(\ce{I2(g)}\). What is the concentration of \(\ce{HI(g)}\) in this mixture?

## Setting Up an Equilibrium Calculation

Very often we use \(K_c\) as a basis for calculating amounts of reactants and products present at equilibrium, starting from some initial concentrations. Alternately, as a means of calculating the value of \(K_c\) for the reaction, we may use information about how initial reactant and product concentrations change to become their values at equilibrium. In either case, it is generally helpful to set up a table, listing initial concentrations, changes in concentrations to reach equilibrium, and values of concentrations at equilibrium. This is sometimes called an ICE table; i.e., initial, change, equilibrium. Depending on the nature of the problem, some of the entries in this kind of table might be specific numbers, while others may need to be algebraic expressions, based on the stoichiometry of the reaction. Regardless of the problem, the entries are always based on the stoichiometric relationships between species in the balance equilibrium equation.

### Q6

Consider the equilibrium

\[\ce{HF(aq) + H2O(l) <=>H3O+ (aq) + F^{–} (aq)} \nonumber \]

and its \(K_c\) expression as you defined it in Q2e. A solution of hydrofluoric acid is prepared by dissolving 0.100 mol \(HF\) in enough water to make a liter of solution. Once equilibrium is established, the concentration of hydronium ion is found to be \([\ce{H3O^{+}} ] = 7.91 \times 10^{–3}\, M\).

- What is the concentration of undissociated HF in the solution? To answer this question, fill in the values in the following ICE table.

ICE Table |
\(\ce{HF}\) | \(\ce{H3O}\) | \(\ce{F^{-}}\) |
---|---|---|---|

Initial |
\(0.100 \,M\) | \(~0\, M\) | \(0 \,M\) |

Change |
|||

Equilibrium |
\(7.91 \times 10^{–3} M\) |

- What is the value of \(K_c\) for \(\ce{HF}\)?
- Based on your value of \(K_c\), is \(\ce{HF}\) a strong or weak acid?
- What is the concentration of undissociated \(\ce{HF}\) in a hydrofluoric acid solution in which \([\ce{H3O^{+}}] = [\ce{F^{–}}] = 0.0100\, M\)?

### Q7

A mixture of 0.297 mol of \(\ce{NO}\) and 0.297 mol of \(\ce{H2}\) is placed in an evacuated 1.00-L vessel at 300 K. The following equilibrium is established:

\[\ce{2 NO(g) + 2 H2(g) <=>N2(g) + 2 H2O(g)} \nonumber \]

At equilibrium, \([\ce{N2}] = 0.120 M\).

- Calculate the equilibrium concentrations of \(\ce{NO}\), \(\ce{H2}\), and \(\ce{H2O}\).
*[Hint: Make a table similar to that shown in Q4, above.]* - Calculate \(K_c\).
- In general, does an equilibrium mixture of \(\ce{NO}\), \(\ce{H2}\), \(\ce{N2}\), and \(\ce{H2O}\) at 300 K favor products or reactants?

## Defining \(K\) in Pressure Units

We have been discussing the equilibrium constant \(K_c\), in which the concentrations are expressed in units of mol/L. If all components that would appear in the expression for the equilibrium constant are gases at constant temperature and volume, it may be more convenient to express the amounts of reactants and products in terms of their partial pressures. From \(PV = nRT\), we can see that individual gas pressures, \(p_i\), in a mixture are proportional to their numbers of moles, \(n_i\), if volume and temperature are held constant; i.e., \(p_i = n_i RT/V\).

Consider the Haber process at equilibrium:

\[\ce{N2(g) + 3 H2(g) <=>2 NH3(g) } \label{HB} \]

We could define the equilibrium constant for this gas-phase reaction either in terms of concentrations (\(K_c\)) or in terms of pressures (\(K_p\)). If chose units of mol/L we would define \(K_c\) as

\[ K_c = \dfrac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3} \nonumber \]

In terms of partial pressures, we could define the equilibrium constant \(K_p\) as

\[ K_p = \dfrac{(p_{\ce{NH3}})^2}{(p_{\ce{N2}})(p_{\ce{H2}})^3} \nonumber \]

Will \(K_p\) and \(K_c\) have the same numerical value for this reaction? Except in special cases, the answer is “no.” To see why there is a difference, let us convert pressures in \(K_p\) to moles per liter using the relationship (from Ideal Gas Law and Dalton's Law)

\[p_i = \dfrac{n_i RT}{V} = \left( \dfrac{n_i}{V} \right) RT \nonumber \]

Recognizing that \(n_i/V\) is mol/L, for each component, for any one component, say species \(A\), we have an expression of the form

\[p_A = \left(\dfrac{n_A}{V}\right)RT = [A]RT \nonumber \]

Substituting into the expression for \(K_p\) for Equation \ref{HB}, we obtain

\[\begin{align} K_p &= \dfrac{([\ce{NH3}]RT)^2}{([\ce{N2}]RT)([\ce{H2}]RT)^3} \\ &= \underbrace{\dfrac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3}}_{K_c} (RT)^{-2} \end{align} \nonumber \]

The first part of this is just \(K_c\), so we may write

\[K_p = K_c(RT)^{-2} \nonumber \]

Notice that the exponent on the \(RT\) term is the difference between the sum of coefficients on gas products (2) minus the sum of coefficients on gas reactants (1 + 3); i.e.,

\[Δn = (2) – (1 + 3) = –2 \nonumber \]

As this specific example shows, the general relationship between \(K_c\) and \(K_p\) is expressed by the equation

\[\boxed{K_p = K_c(RT)^{Δn}} \nonumber \]

where \(Δn\) is the sum of the coefficients of all product gas species minus the sum of the coefficients of all reactant gas species; i.e.,

\[Δn = \sum n_p – \sum n_r \nonumber \]

The following examples illustrate:

Reaction |
\(Δn\) |
\(K_p\) |
---|---|---|

\(\ce{2 NO2(g) <=>N2O4(g) }\) | Δn = 1 – 2 = –1 | \(K_p = K_c(RT)^{–1} \) |

\(\ce{H2(g) + I2(g) <=>2 HI(g)}\) | Δn = 2 – 2 = 0 | \(K_p = K_c\) |

\(\ce{CaO(s) + CO2(g) <=>CaCO3(s) }\) | Δn = 0 – 1 = –1 | \(K_p = K_c(RT)^{–1}\) |

Notice in the second example that \(K_p = K_c\). This will only be true when the sums of gas coefficients for reactants and products are the same. Otherwise, \(Kp \neq K_c\). In the last example, notice that only coefficients of gas species are counted in calculating \(Δn\). The solids are not in the defining expressions of either \(K_p\) or \(K_c\), and so they are ignored in calculating \(Δn\).

### Q8

At 1000 K, the value of \(K_c\) is \(4.08 \times 10^{–3}\) for the equilibrium

\[\ce{2 SO3(g) <=>2 SO2(g) + O2(g)} \nonumber \]

What is the value of \(K_p\) for this reaction under this conditions?

### Q9

At 800 K, the value of \(K_p\) is 4.55 atm^{–1} for the equilibrium

\[\ce{CaO(s) + CO2(g) <=>CaCO3(s)} \nonumber \]

What is the value of \(K_c\) for this reaction under this conditions?