Question

The following equilibria were attained at \(823 \mathrm{~K}\) : $$ \begin{array}{ll} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) & K_{c}=67 \\ \mathrm{CoO}(s)+\mathrm{CO}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) & K_{c}=490 \end{array} $$ Based on these equilibria, calculate the equilibrium constant for \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at \(823 \mathrm{~K}\)

Short answer

The equilibrium constant for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at \(823 \mathrm{~K}\) is \(K_{c} = \frac{67}{490}\).

Step-by-step solution

(Step 1: Write down the given reactions with their equilibrium constants)

Given reactions are: \( \mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \) with \(K_{c1} = 67\) \( \mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) \) with \(K_{c2} = 490\) We need to find the equilibrium constant for the reaction: \( \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \) at \(823 \mathrm{~K}\)

(Step 2: Manipulate given reactions to form the desired reaction)

First, reverse the second given reaction: \( \mathrm{Co}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CoO}(s)+\mathrm{CO}(g) \), with equilibrium constant \(K_{c2}' = \frac{1}{490}\) (When a reaction is reversed, the equilibrium constant is the reciprocal of the initial one.) Now add this reversed reaction and the first given reaction: \( \cancel{\mathrm{CoO}}(s)+\mathrm{H}_{2}(g) + \mathrm{Co}(s)+\cancel{\mathrm{CO}_{2}}(g) \rightleftharpoons \cancel{\mathrm{Co}}(s)+\mathrm{H}_{2} \mathrm{O}(g) + \cancel{\mathrm{CoO}}(s)+\mathrm{CO}(g) \) Simplifying the reaction, we get: \( \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \)

(Step 3: Combine the equilibrium constants)

As we added the reversed second given reaction and the first given reaction to form the desired reaction, we need to multiply their equilibrium constants. So, for the desired reaction, the equilibrium constant \(K_{c}\) is: \(K_{c} = K_{c1} \times K_{c2}' = 67 \times \frac{1}{490} = \frac{67}{490}\)

(Result)

The equilibrium constant for the desired reaction is \(K_{c} = \frac{67}{490}\) at \(823 \mathrm{~K}\).

Key concepts

Equilibrium Constant

The equilibrium constant, often symbolized as \( K_c \), is a vital part of chemical equilibrium. It provides a measure of the relative concentrations of products and reactants present in a chemical reaction at equilibrium under a set temperature. For any reversible reaction at a given temperature:
\[ aA + bB \rightleftharpoons cC + dD \]The equilibrium constant expression \( K_c \) is written as:
\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]This expression depends on the stoichiometry of the reaction, with squared brackets representing the molar concentrations of each species.

• If \( K_c \) is large (greater than one), the equilibrium position favors the formation of products.
• If \( K_c \) is small (less than one), the equilibrium favors reactants.

For the reactions provided in the exercise, each has a particular \( K_c \):
- The reaction involving \( \mathrm{H}_2 \) and \( \mathrm{H}_2\mathrm{O} \) has \( K_{c1} = 67 \).
- The reaction between \( \mathrm{CO} \) and \( \mathrm{CO}_2 \) has \( K_{c2} = 490 \).These values help us understand the moderate and high tendencies towards the formation of respective products in these reactions at 823 K.

Reaction Manipulation

Sometimes, reactions must be modified to determine the equilibrium constant for another related reaction. This involves reversing reactions or combining them, changing the equilibrium expressions accordingly.
When a reaction is reversed, the new equilibrium constant \( K'_c \) is the reciprocal of the original, such that \( K'_c = 1/K_c \).
For example, reversing the reaction:
\[ \mathrm{CoO}(s) + \mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s) + \mathrm{CO}_2(g) \] changes \( K_{c2} = 490 \) to \( K'_{c2} = 1/490 \).

• Reversing changes directions and also the position of equilibrium.
• The equilibrium constant flips and can drastically affect the speed to reactance or product side.

If two reactions are added together to form a new reaction, their equilibrium constants are multiplied. When trying to find the equilibrium constant of the reaction:
\( \mathrm{H}_{2}(g) + \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H}_{2}\mathrm{O}(g) \)the given reactions were combined, and the equilibrium constants multiplied as shown:
\[ K_c = K_{c1} \times \frac{1}{K_{c2}} \] This correctly finds an equivalent \( K_c \) for the newly formulated reaction.

Equilibrium Calculations

Calculating the equilibrium constant for any derived reaction is a step-by-step process where known reactions are manipulated and combined. First, note the reactions and their \( K_c \) values. Reverse any necessary reactions to reflect desired participants and results, adjusting \( K_c \) accordingly.
In boiling this down for the new reaction:1. Reverse the second reaction in the problem statement.2. The new reaction becomes multiplied by the first reaction, aligning products and reactants.3. The resulting \( K_c \) calculation combines the altered constants.
In our example:The equilibrium constant sought is:\[ K_{c} = 67 \times \frac{1}{490} = \frac{67}{490} \]

• Each step builds onto the previous.
• This ensures transformations are correct and maintain correct kinetic properties.

Remember, minor calculation errors can mislead conclusions about equilibrium positions, so each step should be approached with care.