13.3: Equilibrium Constants
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Learning Objectives
By the end of this section, you will be able to:
- Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions
- Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
- Relate the magnitude of an equilibrium constant to properties of the chemical system
The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient ( Q ) . For a reversible reaction described by
\[m A +n B +\rightleftharpoons x C +y D \nonumber \]
the reaction quotient is derived directly from the stoichiometry of the balanced equation as
\[Q_c=\frac{[ C ]^x[ D ]^y}{[ A ]^m[ B ]^n} \nonumber \]
where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:
\[Q_p=\frac{P_{ C }^x P_{ D }^y}{P_{ A }{ }^m P_{ B }{ }^n} \nonumber \]
Note that the reaction quotient equations above are a simplification of more rigorous expressions that use relative values for concentrations and pressures rather than absolute values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing Q . In most cases, this will introduce only modest errors in calculations involving reaction quotients.
Example \(\PageIndex{1}\): Writing Reaction Quotient Expressions
Write the concentration-based reaction quotient expression for each of the following reactions:
- \(\ce{3 O2(g) \rightleftharpoons 2 O3(g)}\)
- \(\ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)}\)
- \(\ce{4 NH3(g) + 7 O2(g) \rightleftharpoons 4 NO2(g) + 6 H2O(g)}\)
Solution
- \[Q_c=\frac{\left[ \ce{O3} \right]^2}{\left[ \ce{O2} \right]^3} \nonumber \]
- \[Q_c=\frac{\left[ \ce{NH3} \right]^2}{\left[ \ce{N2} \right]\left[ \ce{H2} \right]^3} \nonumber \]
- \[Q_c=\frac{\left[ \ce{NO2} \right]^4\left[ \ce{H2O} \right]^6}{\left[ \ce{NH3} \right]^4\left[ \ce{O2} \right]^7} \nonumber \]
Exercise \(\PageIndex{1}\)
Write the concentration-based reaction quotient expression for each of the following reactions:
- \(\ce{2SO2(g) + O2(g) \rightleftharpoons 2 SO3(g)}\)
- \(\ce{C4H8(g) \rightleftharpoons 2 C2H4(g)}\)
- \(\ce{2 C4H10(g) + 13 O2(g) \rightleftharpoons 8 CO2(g) + 10 H2O (g)}\)
- Answer
-
- \[Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]} \nonumber \]
- \[Q_c=\frac{\left[ \ce{C2H4} \right]^2}{\left[ \ce{C4H8} \right]} \nonumber \]
- \[Q_c=\frac{\left[ \ce{CO2} \right]^8\left[ \ce{H2O} \right]^{10}}{\left[ \ce{C4H10} \right]^2\left[ \ce{O2} \right]^{13}} \nonumber \]
The numerical value of \(Q\) varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction’s status. To illustrate this point, consider the oxidation of sulfur dioxide:
\[\ce{2 SO2(g) + O2(g) \rightleftharpoons 2 SO3(g)} \nonumber \]
Two different experimental scenarios are depicted in Figure \(\PageIndex{1}\), one in which this reaction is initiated with a mixture of reactants only, SO 2 and O 2 , and another that begins with only product, SO 3 . For the reaction that begins with a mixture of reactants only, Q is initially equal to zero:
\[Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=\frac{0^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=0 \nonumber \]
As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of Q c ), product concentration increases (as does the numerator of Q c ), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of Q c .
If the reaction begins with only product present, the value of Q c is initially undefined (immeasurably large, or infinite):
\[Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=\frac{\left[ \ce{SO3} \right]^2}{0} \rightarrow \infty \nonumber \]
In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of Q c decrease with time, the reactant concentrations and the denominator of Q c increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.
The constant value of Q exhibited by a system at equilibrium is called the equilibrium con stant, \(K\):
\[K \equiv Q \text { at equilibrium } \nonumber \]
Comparison of the data plots in Figure \(\PageIndex{1}\) shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the law of mass action .
Definition: Law of Mass Action
At a given temperature, the reaction quotient for a system at equilibrium is constant.
Example \(\PageIndex{2}\): Evaluating a Reaction Quotient
Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:
\[\ce{2 NO2(g) <=> N2O4(g)} \nonumber \]
When 0.10 mol NO 2 is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO 2 ] = 0.016 M and [N 2 O 4 ] = 0.042 M .
- What is the value of the reaction quotient before any reaction occurs?
- What is the value of the equilibrium constant for the reaction?
Solution
As for all equilibrium calculations in this text, use the simplified equations for \(Q\) and \(K\) and disregard any concentration or pressure units, as noted previously in this section.
(a) Before any product is formed
\[\left[ \ce{NO2} \right]=\frac{0.10~\text{mol} }{1.0~\text{L} }=0.10~\text{M} \nonumber \]
\[[\ce{N2O4}] = 0~\text{M} \nonumber \]
Thus
\[Q_c=\frac{\left[ \ce{N2O4} \right]}{\left[ \ce{NO2} \right]^2}=\frac{0}{0.10^2}=0 \nonumber \]
(b) At equilibrium,
\[K_c=Q_c=\frac{\left[ \ce{N2O4} \right]}{\left[ \ce{NO2} \right]^2}=\frac{0.042}{0.016^2}=1.6 \times 10^2. \nonumber \]
The equilibrium constant is \(1.6 \times 10^{2}\).
Exercise \(\PageIndex{2}\)
For the reaction
\[\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)} \nonumber \]
the equilibrium concentrations are [SO 2 ] = 0.90 M , [O 2 ] = 0.35 M , and [SO 3 ] = 1.1 M . What is the value of the equilibrium constant, K c ?
- Answer
-
K c = 4.3
By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large K will reach equilibrium when most of the reactant has been converted to product, whereas a small K indicates the reaction achieves equilibrium after very little reactant has been converted. It’s important to keep in mind that the magnitude of K does not indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.
The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing Q to K for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur.
To further illustrate this important point, consider the reversible reaction shown below:
\[\ce{CO(g) + H2O(g) \rightleftharpoons CO2(g) + H2(g)} \quad K_c=0.640 \quad T =800{ }^{\circ} C \nonumber \]
The bar charts in Figure \(\PageIndex{2}\) represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction’s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established.
Example \(\PageIndex{3}\): Predicting the Direction of Reaction
Given here are the starting concentrations of reactants and products for three experiments involving this reaction:
\[\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)} \quad K_c=0.64 \nonumber \]
Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.
| Reactants/Products | Experiment 1 | Experiment 2 | Experiment 3 |
|---|---|---|---|
| [CO] i | 0.020 M | 0.011 M | 0.0094 M |
| [H 2 O] i | 0.020 M | 0.0011 M | 0.0025 M |
| [CO 2 ] i | 0.0040 M | 0.037 M | 0.0015 M |
| [H 2 ] i | 0.0040 M | 0.046 M | 0.0076 M |
Solution
Experiment 1:
\[Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.0040)(0.0040)}{(0.020)(0.020)}=0.040 \nonumber \]
Q c < K c (0.040 < 0.64)
The reaction will proceed in the forward direction.
Experiment 2:
\[Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.037)(0.046)}{(0.011)(0.0011)}=1.4 \times 10^2 \nonumber \]
Q c > K c (140 > 0.64)
The reaction will proceed in the reverse direction.
Experiment 3:
\[Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.0015)(0.0076)}{(0.0094)(0.0025)}=0.48 \nonumber \]
Q c < K c (0.48 < 0.64)
The reaction will proceed in the forward direction.
Exercise \(\PageIndex{3}\)
Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.
- A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl 2 (g), and 0.500 mol of NOCl: \[\ce{2 NO(g) + Cl2(g) <=> 2 NOCl(g)} \quad K_c=4.6 \times 10^4 \nonumber \]
- A 5.0-L flask containing 17 g of NH 3 , 14 g of N 2 , and 12 g of H 2 : \[\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)} \quad K_c=0.060 \nonumber \]
- A 2.00-L flask containing 230 g of SO 3 (g): \[\ce{2 SO3(g) <=> 2 SO2(g) + O2(g)} \quad K_c=0.230 \nonumber \]
- Answer
-
(a) Q c = 6.45 \times 10^{3} DELMAR, forward. (b) Q c = 0.23, reverse. (c) Q c = 0, forward.
Homogeneous Equilibria
A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions . These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:
\[\begin{aligned}
\ce{C2H2(aq) + 2 Br2(aq) & \rightleftharpoons C2H2Br4(aq)} & K_c & =\frac{\left[ C_2 H_2 Br_4\right]}{\left[ C_2 H_2\right]\left[ Br_2\right]^2} \\[4pt]
\ce{I2(aq) + I^{-}(aq) & \rightleftharpoons I_3^{-}(aq)} & K_c & =\frac{\left[ I_3-\right.}{\left[ I_2\right]\left[ I^{-}\right]} \\[4pt]
\ce{HF(aq) + H2O(l) & \rightleftharpoons H3O^{+}(aq) + F^{-}(aq)} & K_c & =\frac{\left[ H_3 O^{+}\right]\left[ F^{-}\right]}{[ HF ]} \\[4pt]
\ce{NH3(aq) + H2O(l) & \rightleftharpoons NH4^{+}(aq) + OH^{-}(aq)} & K_c & =\frac{\left[ NH_4^{+}\right]\left[ OH^{-}\right]}{\left[ NH_3\right]}
\end{aligned} \nonumber \]
These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the Q (or K ) expression mentioned previously in this chapter, in which relative concentrations for liquids and solids are equal to 1 and needn’t be included . Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.
Note
It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that K a = K eq [H 2 O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation K a = K eq (\(\textit{a}_{H_2O}\)). Because \(\textit{a}_{H_2O}\) = 1 for a dilute solution, K a = K eq (1), or K a = K eq .
The equilibria below all involve gas-phase solutions:
\[\begin{aligned}
\ce{C2H6(g) & \rightleftharpoons C2H4(g) + H2(g)} & K_c & =\frac{\left[ C_2 H_4\right]\left[ H_2\right]}{\left[ C_2 H_6\right]} \\[4pt]
\ce{3 O2(g) & \rightleftharpoons 2 O3(g)} & K_c & =\frac{\left[ O_3\right]^2}{\left[ O_2\right]^3} \\[4pt]
\ce{N2(g) + 3 H2(g) & \rightleftharpoons 2 NH3(g)} & K_c & =\frac{\left[ NH_3\right]^2}{\left[ N_2\right]\left[ H_2\right]^3} \\[4pt]
\ce{C3H8(g) + 5 O2(g) & \rightleftharpoons 3 CO2(g) + 4 H2O(g)} & K_c & =\frac{\left[ CO_2\right]^3\left[ H_2 O \right]^4}{\left[ C_3 H_8\right]\left[ O_2\right]^5}
\end{aligned} \nonumber \]
For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations ( K c ) or partial pressures ( K p ) of the reactants and products. A relation between these two K values may be simply derived from the ideal gas equation and the definition of molarity:
\[\begin{align*}
P V &=n R T \\[4pt]
P & =\left(\frac{n}{V}\right) R T \\[4pt]
& =M R T
\end{align*} \nonumber \]
where \(P\) is partial pressure, \(V\) is volume, \(n\) is molar amount, \(R\) is the gas constant, \(T\) is temperature, and \(M\) is molar concentration.
For the gas-phase reaction
\[a A +b B \rightleftharpoons c C +d D \nonumber \]
\[\begin{align*}
K_P &=\frac{\left(P_C\right)^c\left(P_D\right)^d}{\left(P_A\right)^a\left(P_B\right)^b} \\[4pt]
&= \dfrac{([ C ] \times R T)^c([ D ] \times R T)^d}{([ A ] \times R T)^a([ B ] \times R T)^b} \\[4pt]
&= \dfrac{[ C ]^c[ D ]^d}{[ A ]^a[ B ]^b} \times \frac{(R T)^{c+d}}{(R T)^{a+b}} \\[4pt]
&= K_c(R T)^{(c+d)-(a+b)} \\[4pt]
&= K_c(R T)^{\Delta n}
\end{align*} \nonumber \]
And so, the relationship between K c and K P is
\[K_P=K_c(R T)^{\Delta n} \nonumber \]
where \(Δn\) is the difference in the molar amounts of product and reactant gases, in this case:
\[\Delta n=(c+d)-(a+b) \nonumber \]
Example \(\PageIndex{4}\): Calculation of K P
Write the equations relating K c to K P for each of the following reactions:
- \(\ce{C2H6(g) <=> C2H4(g) + H2(g)}\)
- \(\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}\)
- \(\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}\)
- K c is equal to 0.28 for the following reaction at 900 °C: \[\ce{CS2(g) + 4 H2(g) <=> CH4(g) + 2 H2S(g)} \nonumber \] What is K P at this temperature?
Solution
- \(Δn = (2) − (1) = 1\) \[K_P = K_c (RT)^{Δn} = K_c (RT)^1 = K_c (RT) \nonumber \]
- \(Δn = (2) − (2) = 0\) \[K_P = K_c (RT)^{Δn} = K_c (RT)^0 = K_c \nonumber \]
- \(Δn = (2) − (1 + 3) = −2\) \[K_P = K_c (RT)^{Δn} = K_c (RT)^{−2} = \dfrac{K_c}{(R T)^2} \nonumber \]
- \[K_P = K_c (RT)^{Δn} = (0.28)[(0.0821)(1173)]^{−2} = 3.0 \times 10^{−5} \nonumber \]
Exercise \(\PageIndex{4}\)
Write the equations relating K c to K P for each of the following reactions:
- \(\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)}\)
- \(\ce{N2O4(g) <=> 2 NO2(g)}\)
- \(\ce{C3H8(g) + 5 O2(g) <=> 3 CO2(g) + 4 H2O (g)}\)
- At 227 °C, the following reaction has K c = 0.0952: \[\ce{CH3OH(g) <=> CO(g) + 2 H2(g)} \nonumber \] What would be the value of K P at this temperature?
- Answer
-
(a) K P = K c ( RT ) −1 ; (b) K P = K c ( RT ); (c) K P = K c ( RT ); (d) 160 or 1.6 \times 10^{2} DELMAR
Heterogeneous Equilibria
A heterogeneous equilibrium involves reactants and products in two or more different phases, as illustrated by the following examples:
\[\begin{align*}
\ce{PbCl2(s) & \rightleftharpoons Pb^{2+}(aq) + 2 Cl^{-}(aq)} & K_c & =\left[ \ce{Pb^{2+}} \right]\left[ \ce{Cl^{-}} \right]^2 \\[4pt]
\ce{CaO(s) + CO2(g) & \rightleftharpoons CaCO3(s)} & K_c & =\frac{1}{\left[ \ce{CO_2}\right]} \\[4pt]
\ce{C(s) + 2S(g) & \rightleftharpoons CS2(g)} & K_c & =\frac{\left[ CS_2\right]}{\left[ \ce{S^2} \right.} \\[4pt]
\ce{Br2(l) & \rightleftharpoons Br2(g)} & K_c & =\left[ \ce{Br_2(g)} \right]
\end{align*} \nonumber \]
Again, note that concentration terms are only included for gaseous and solute species, as discussed previously.
Two of the above examples include terms for gaseous species only in their equilibrium constants, and so K p expressions may also be written:
\[\begin{align*}
\ce{CaO(s) + CO2(g) & \rightleftharpoons CaCO3(s)} & K_P & =\frac{1}{P_{ \ce{CO2}}} \\[4pt]
\ce{C(s) + 2 S(g) & \rightleftharpoons CS2(g)} & K_P & =\frac{P_{ \ce{CS2}}}{\left(P_{ \ce{S} }\right)^2}
\end{align*} \nonumber \]
Coupled Equilibria
The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more coupled equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.
1. Changing the direction of a chemical equation essentially swaps the identities of “reactants” and “products,” and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation.
\[\begin{array}{ll}
A \rightleftharpoons B & K_{ c }=\frac{[ B ]}{[ A ]} \\[4pt]
B \rightleftharpoons A & K_{ c^{\prime}}=\frac{[ A ]}{[ B ]}
\end{array} \nonumber \]
\[K_{ c^{\prime}}=\frac{1}{ K_{ c }} \nonumber \]
2. Changing the stoichiometric coefficients in an equation by some factor x results in an exponential change in the equilibrium constant by that same factor:
\[\begin{array}{ll}
A \rightleftharpoons B & K_{ c }=\frac{[ B ]}{[ A ]} \\[4pt]
xA \rightleftharpoons xB & K_{ c }=\frac{[ B ]^{ x }}{[ A ]^{ x }}
\end{array} \nonumber \]
\[K_{ c^{\prime}}= K_{ c }{ }^{ x } \nonumber \]
3. Adding two or more equilibrium equations together yields an overall equation whose equilibrium constant is the mathematical product of the individual reaction’s K values:
\[\begin{array}{ll}
A \rightleftharpoons B & K_{ c 1}=\frac{[ B ]}{[ A ]} \\[4pt]
B \rightleftharpoons C & K_{ c 2}=\frac{[ C ]}{[ B ]}
\end{array} \nonumber \]
The net reaction for these coupled equilibria is obtained by summing the two equilibrium equations and canceling any redundancies:
\[\begin{aligned}
& A + B \rightleftharpoons B + C \\[4pt]
& A + B \rightleftharpoons B + C \\[4pt]
& A \rightleftharpoons C
\end{aligned} \nonumber \]
\[K_{ c^{\prime}}=\frac{[ C ]}{[ A ]} \nonumber \]
Comparing the equilibrium constant for the net reaction to those for the two coupled equilibrium reactions reveals the following relationship:
\[K_{ c 1} K_{ c 2}=\frac{[ B ]}{[ A ]} \times \frac{[ C ]}{[ B ]}=\frac{ \cancel{[ B ]}[ C ]}{[ A ] \cancel{[ B ]}}=\frac{[ C ]}{[ A ]}= K_{ c^{\prime}} \nonumber \]
\[K_{ c^{\prime}}= K_{ c 1} K_{ c 2} \nonumber \]
Example \(\PageIndex{5}\) demonstrates the use of this strategy in describing coupled equilibrium processes.
Example \(\PageIndex{5}\): Equilibrium Constants for Coupled Reactions
A mixture containing nitrogen, hydrogen, and iodine established the following equilibrium at 400 °C:
\[\ce{2 NH3(g) + 3 I2(g) \rightleftharpoons N2(g) + 6 HI(g)} \nonumber \]
Use the information below to calculate Kc for this reaction.
\[\begin{align*}
\ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)} & K_{ c 1}=0.50 \text { at } 400{ }^{\circ} C \\[4pt]
\ce{H2(g) + I2(g) \rightleftharpoons 2 HI(g)} & K_{ c 2}=50 \text { at } 400{ }^{\circ} C
\end{align*} \nonumber \]
Solution
The equilibrium equation of interest and its K value may be derived from the equations for the two coupled reactions as follows.
Reverse the first coupled reaction equation:
\[2 NH_3(g) \rightleftharpoons N_2(g) + 3 H_2(g) \quad K_{ c 1}{ }^{\prime}=\frac{1}{ K_{ c 1}}=\frac{1}{0.50}=2.0 \nonumber \]
Multiply the second coupled reaction by 3:
\[3 H_2(g) + 3 I_2(g) \rightleftharpoons 6 HI (g) \quad K_{ c 2}{ }^{\prime}= K_{ c 2}^3=50^3=1.2 \times 10^5 \nonumber \]
Finally, add the two revised equations:
\[\begin{align*}
\ce{2 NH3(g) + 3 H2(g) + 3 I2(g) &\rightleftharpoons N2(g) + 3 H2(g) + 6 HI(g)} \\[4pt]
\ce{2 NH3(g) + 3 I2(g) &\rightleftharpoons N2(g) + 6HI(g)}
\end{align*} \nonumber \]
\[K_{ c }= K_{ c 1}, K_{ c 2},=(2.0)\left(1.2 \times 10^5\right)=2.5 \times 10^5 \nonumber \]
Exercise \(\PageIndex{5}\)
Use the provided information to calculate Kc for the following reaction at 550 °C:
\[\begin{align*}
\ce{H2(g) + CO2(g) &\rightleftharpoons CO(g) + H2O(g)} && K_{ c }=? \\[4pt]
\ce{CoO(s) + CO(g) &\rightleftharpoons Co(s)+ CO2(g)} && K_{ c 1}=490 \\[4pt]
\ce{CoO(s) + H2(g) &\rightleftharpoons Co(s)+ H2O(g)} && K_{ c 1}=67
\end{align*} \nonumber \]
- Answer
-
K c = 0.14