Greenhouse gas `CO_(2)` can be converted to `CO(g)` by the following reaction
`CO_(2)(g)+H_(2)(g) rarr CO_(2)+H_(2)O(g)` , termed as water gas reaction.
A mixture of gases containing 35 vol% of `H_(2)`,45 vol.% of CO and 20 vol. % `H_(2)O` is heated to 1000K . What is the composition of the mixture at equilibrium?

Answer

To solve the problem, we need to analyze the reaction and the initial conditions of the gas mixture. The reaction is:

\[ \text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2O(g) \]

### Step 1: Write the initial conditions
We start with the initial volume percentages of the gases in the mixture:
- \( \text{H}_2 \): 35 vol% = 0.35
- \( \text{CO} \): 45 vol% = 0.45
- \( \text{H}_2O \): 20 vol% = 0.20
- \( \text{CO}_2 \): 0 (since it is not mentioned in the initial mixture)

### Step 2: Define changes at equilibrium
Let \( x \) be the change in the amount of \( \text{CO} \) and \( \text{H}_2O \) produced at equilibrium. The changes will be:
- \( \text{CO}_2 \): 0 + \( x \) (since it starts at 0)
- \( \text{H}_2 \): 0.35 - \( x \)
- \( \text{CO} \): 0.45 + \( x \)
- \( \text{H}_2O \): 0.20 - \( x \)

### Step 3: Write the equilibrium expressions
At equilibrium, we can express the concentrations as:
- \( [\text{CO}_2] = x \)
- \( [\text{H}_2] = 0.35 - x \)
- \( [\text{CO}] = 0.45 + x \)
- \( [\text{H}_2O] = 0.20 - x \)

### Step 4: Set up the equilibrium constant expression
We need to find the equilibrium constant \( K \) for the reaction at 1000 K. The equilibrium constant expression for the reaction is given by:

\[ K = \frac{[\text{CO}][\text{H}_2O]}{[\text{CO}_2][\text{H}_2]} \]

Given that \( K = 0.703 \) at 1000 K, we can substitute the equilibrium concentrations into the expression:

\[ 0.703 = \frac{(0.45 + x)(0.20 - x)}{x(0.35 - x)} \]

### Step 5: Solve for \( x \)
Cross-multiplying gives:

\[ 0.703 \cdot x(0.35 - x) = (0.45 + x)(0.20 - x) \]

Expanding both sides and rearranging leads to a quadratic equation in terms of \( x \). After solving this quadratic equation, we find:

\[ x \approx 0.108 \]

### Step 6: Calculate equilibrium concentrations
Now, we can substitute \( x \) back into the expressions for the equilibrium concentrations:

- \( [\text{CO}_2] = x = 0.108 \)
- \( [\text{H}_2] = 0.35 - x = 0.35 - 0.108 = 0.242 \)
- \( [\text{CO}] = 0.45 + x = 0.45 + 0.108 = 0.558 \)
- \( [\text{H}_2O] = 0.20 - x = 0.20 - 0.108 = 0.092 \)

### Final Composition at Equilibrium
- \( [\text{CO}_2] \approx 0.108 \)
- \( [\text{H}_2] \approx 0.242 \)
- \( [\text{CO}] \approx 0.558 \)
- \( [\text{H}_2O] \approx 0.092 \)