For the reaction
`H_(2)(g)+CO_(2) (g)hArrCO(g)+H_(2)O(g),` if the initial concentration of`[H_(2)]=[CO_(2)]`and x moles /litres of hydrogen is consumed at equilibrium , the correct expression of `K_(p)` is :

Options

A. (a) `(x^(2))/((1-x)^(2))`
B. (b) `((1-x)^(2))/((1-x)^(2))`
C. (c) `(x^(2))/((2+x)^(2))`
D. (d) `(x^(2))/((1-x)^(2))`

Answer

To solve the problem regarding the equilibrium constant \( K_p \) for the reaction

\[ H_2(g) + CO_2(g) \rightleftharpoons CO(g) + H_2O(g), \]

we will follow these steps:

### Step 1: Define Initial Concentrations
Let the initial concentration of both \( H_2 \) and \( CO_2 \) be \( A \) moles/litre.

### Step 2: Define Changes at Equilibrium
Since \( x \) moles/litre of \( H_2 \) are consumed at equilibrium, the changes in concentration can be defined as follows:
- The concentration of \( H_2 \) at equilibrium will be \( [H_2] = A - x \).
- The concentration of \( CO_2 \) at equilibrium will also be \( [CO_2] = A - x \).
- The concentration of \( CO \) at equilibrium will be \( [CO] = x \).
- The concentration of \( H_2O \) at equilibrium will also be \( [H_2O] = x \).

### Step 3: Write the Expression for \( K_c \)
The expression for the equilibrium constant \( K_c \) in terms of concentrations is given by:

\[ K_c = \frac{[CO][H_2O]}{[H_2][CO_2]}. \]

Substituting the equilibrium concentrations we found:

\[ K_c = \frac{x \cdot x}{(A - x)(A - x)} = \frac{x^2}{(A - x)^2}. \]

### Step 4: Calculate \( \Delta N_g \)
Next, we need to calculate \( \Delta N_g \), which is the change in the number of moles of gas from reactants to products:

- Moles of gaseous products = 1 (from CO) + 1 (from \( H_2O \)) = 2.
- Moles of gaseous reactants = 1 (from \( H_2 \)) + 1 (from \( CO_2 \)) = 2.

Thus,

\[ \Delta N_g = 2 - 2 = 0. \]

### Step 5: Relate \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by:

\[ K_p = K_c \cdot (RT)^{\Delta N_g}. \]

Since \( \Delta N_g = 0 \), we have:

\[ K_p = K_c \cdot (RT)^0 = K_c. \]

### Step 6: Final Expression for \( K_p \)
Thus, we can conclude that:

\[ K_p = \frac{x^2}{(A - x)^2}. \]

### Step 7: Simplifying the Expression
Since \( K_p = K_c \), we can express it as:

\[ K_p = \frac{x^2}{(A - x)^2}. \]

### Conclusion
The correct expression for \( K_p \) is:

\[ K_p = \frac{x^2}{(A - x)^2}. \]