In a 20 litre vessel at 300 K initially, 1 mole of `CO, H_(2)O, CO_(2)` are present, then for the equilibrium of `CO+H_(2)OhArr CO_(2)+H_(2)` which of the following is true?

Options

A. concentration of CO does not change from initial state
B. concentration of CO is constant when equilibrium is achieved
C. `H_(2)O` behave as a reducing agent
D. All of these are true

Answer

To solve the problem, we need to analyze the equilibrium reaction given:

\[ \text{CO} + \text{H}_2\text{O} \rightleftharpoons \text{CO}_2 + \text{H}_2 \]

### Step 1: Determine Initial Concentrations
Initially, we have 1 mole each of CO, H2O, and CO2 in a 20-liter vessel. The initial concentration of each component can be calculated using the formula:

\[ \text{Concentration (M)} = \frac{\text{Number of moles}}{\text{Volume (L)}} \]

- For CO:
\[ \text{[CO]}_0 = \frac{1 \text{ mol}}{20 \text{ L}} = 0.05 \text{ M} \]

- For H2O:
\[ \text{[H}_2\text{O]}_0 = \frac{1 \text{ mol}}{20 \text{ L}} = 0.05 \text{ M} \]

- For CO2:
\[ \text{[CO}_2]_0 = \frac{1 \text{ mol}}{20 \text{ L}} = 0.05 \text{ M} \]

- For H2:
\[ \text{[H}_2]_0 = 0 \text{ M} \quad (\text{since no H}_2 \text{ is present initially}) \]

### Step 2: Set Up the Change in Concentrations
Let \( x \) be the amount of CO and H2O that reacts at equilibrium. The changes in concentration can be expressed as follows:

- For CO:
\[ \text{[CO]} = 0.05 - x \]

- For H2O:
\[ \text{[H}_2\text{O]} = 0.05 - x \]

- For CO2:
\[ \text{[CO}_2] = 0.05 + x \]

- For H2:
\[ \text{[H}_2] = 0 + x \]

### Step 3: Analyze the Statements
Now we will analyze the statements given in the question:

1. **Concentration of CO does not change from initial state.**
- This statement is **false** because the concentration of CO changes as it reacts to form products.

2. **Concentration of CO is constant when equilibrium is achieved.**
- This statement is also **false** because the concentration of CO will change until equilibrium is reached.

3. **H2O behaves as a reducing agent.**
- This statement is **false** because H2O is being oxidized (losing electrons) in this reaction, thus it behaves as an oxidizing agent, not a reducing agent.

### Conclusion
Since all three statements are incorrect, the answer to the question is that none of the statements are true regarding the equilibrium of the reaction.