At a certain temperature the following equilibrium is established `CO(g)+NO_2(g)hArrCO_2(g)+NO(g)`
One mole of each of the four gas is mixed in one litre container and the reaction is allowed to reach equilibrium state.When excess of baryta water `(Ba(OH)_2)` is added to the equilibrium mixture, the weight of white ppt. `(BaCO_3)` obtained is 236.4 gm.The equilibrium constant `K_C` of the reaction is (Ba=137)

Options

A. 1.2
B. 2.25
C. 2.1
D. 3.6

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the equilibrium expression
The equilibrium reaction is:
\[ \text{CO(g)} + \text{NO}_2(g) \rightleftharpoons \text{CO}_2(g) + \text{NO(g)} \]

### Step 2: Set up the initial conditions
Initially, we have 1 mole of each gas in a 1-liter container:
- [CO] = 1 M
- [NO2] = 1 M
- [CO2] = 0 M
- [NO] = 0 M

### Step 3: Define the change in concentration
Let \( x \) be the amount of CO that reacts at equilibrium. Thus, at equilibrium:
- [CO] = \( 1 - x \)
- [NO2] = \( 1 - x \)
- [CO2] = \( x \)
- [NO] = \( x \)

### Step 4: Calculate the amount of BaCO3 produced
When excess baryta water \( \text{Ba(OH)}_2 \) is added, it reacts with \( \text{CO}_2 \) to form \( \text{BaCO}_3 \):
\[ \text{CO}_2 + \text{Ba(OH)}_2 \rightarrow \text{BaCO}_3 + \text{H}_2O \]

Given that the weight of \( \text{BaCO}_3 \) formed is 236.4 g, we can calculate the number of moles of \( \text{BaCO}_3 \):
- Molar mass of \( \text{BaCO}_3 \) = 137 (Ba) + 12 (C) + 3 × 16 (O) = 197 g/mol
- Moles of \( \text{BaCO}_3 \) = \( \frac{236.4 \, \text{g}}{197 \, \text{g/mol}} = 1.2 \, \text{mol} \)

### Step 5: Relate moles of \( \text{CO}_2 \) to \( x \)
Since 1 mole of \( \text{CO}_2 \) produces 1 mole of \( \text{BaCO}_3 \), we have:
\[ x = 1.2 \]

### Step 6: Substitute \( x \) back to find concentrations
Now substituting \( x \) back:
- [CO] = \( 1 - x = 1 - 1.2 = -0.2 \) (This indicates that the reaction goes to completion)
- [NO2] = \( 1 - x = 1 - 1.2 = -0.2 \) (This indicates that the reaction goes to completion)
- [CO2] = \( x = 1.2 \)
- [NO] = \( x = 1.2 \)

### Step 7: Calculate equilibrium concentrations
However, since we cannot have negative concentrations, we need to consider that the reaction goes to completion:
- At equilibrium, we have:
- [CO] = 0
- [NO2] = 0
- [CO2] = 1.2
- [NO] = 1.2

### Step 8: Write the equilibrium constant expression
The equilibrium constant \( K_C \) is given by:
\[ K_C = \frac{[\text{CO}_2][\text{NO}]}{[\text{CO}][\text{NO}_2]} \]

### Step 9: Substitute the equilibrium concentrations into the expression
Substituting the values:
\[ K_C = \frac{(1.2)(1.2)}{(0)(0)} \]

Since we cannot divide by zero, we need to consider that the reaction goes to completion, and thus we can express \( K_C \) in terms of the initial moles:
\[ K_C = \frac{(1.2)^2}{(1 - 1.2)(1 - 1.2)} \]

### Step 10: Calculate \( K_C \)
Since we are looking for the equilibrium constant when the reaction goes to completion:
\[ K_C = \frac{(1.2)^2}{(0)(0)} \]

This indicates that the equilibrium constant is very large, approaching infinity, indicating that the reaction favors the products.

### Conclusion
Thus, the equilibrium constant \( K_C \) is effectively very large, and we can conclude that the reaction goes to completion.