The equilibrium constant `K_(p)` for the reaction
`H_(2)(g)+CO_(2)(g)hArrH_(2)O(g)+CO(g)`
is 4.0 at `1660^(@)C` Initially `0.80` mole `H_(2) ` and `0.80` mole `CO_(2)` are injected into a 5.0 litre flask. What is the equilibrium concentration of `CO_(2)(g)`?

Options

A. `0.533`
B. `0.0534`
C. `0.535`
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ H_2(g) + CO_2(g) \rightleftharpoons H_2O(g) + CO(g) \]

### Step 2: Determine initial moles and concentrations
Initially, we have:
- Moles of \( H_2 = 0.80 \)
- Moles of \( CO_2 = 0.80 \)
- Moles of \( H_2O = 0 \)
- Moles of \( CO = 0 \)

The volume of the flask is \( 5.0 \) liters. Therefore, the initial concentrations are:
- Concentration of \( H_2 = \frac{0.80}{5.0} = 0.16 \, \text{M} \)
- Concentration of \( CO_2 = \frac{0.80}{5.0} = 0.16 \, \text{M} \)
- Concentration of \( H_2O = 0 \)
- Concentration of \( CO = 0 \)

### Step 3: Set up the change in moles at equilibrium
Let \( x \) be the amount of \( H_2 \) and \( CO_2 \) that reacts at equilibrium. At equilibrium, the moles will be:
- Moles of \( H_2 = 0.80 - x \)
- Moles of \( CO_2 = 0.80 - x \)
- Moles of \( H_2O = x \)
- Moles of \( CO = x \)

### Step 4: Write the equilibrium concentrations
The equilibrium concentrations will be:
- Concentration of \( H_2 = \frac{0.80 - x}{5.0} \)
- Concentration of \( CO_2 = \frac{0.80 - x}{5.0} \)
- Concentration of \( H_2O = \frac{x}{5.0} \)
- Concentration of \( CO = \frac{x}{5.0} \)

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[H_2O][CO]}{[H_2][CO_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{x}{5.0}\right)\left(\frac{x}{5.0}\right)}{\left(\frac{0.80 - x}{5.0}\right)\left(\frac{0.80 - x}{5.0}\right)} = \frac{x^2}{(0.80 - x)^2} \]

### Step 6: Substitute the value of \( K_c \)
Given that \( K_p = 4.0 \) and since \( K_p = K_c \) (as \( \Delta n_g = 0 \)), we have:
\[ 4 = \frac{x^2}{(0.80 - x)^2} \]

### Step 7: Solve for \( x \)
Taking the square root of both sides:
\[ 2 = \frac{x}{0.80 - x} \]
Cross-multiplying gives:
\[ 2(0.80 - x) = x \]
Expanding and rearranging:
\[ 1.6 - 2x = x \implies 1.6 = 3x \implies x = \frac{1.6}{3} \approx 0.533 \]

### Step 8: Calculate the equilibrium concentration of \( CO_2 \)
Now, we can find the equilibrium concentration of \( CO_2 \):
\[ \text{Equilibrium concentration of } CO_2 = \frac{0.80 - x}{5.0} = \frac{0.80 - 0.533}{5.0} = \frac{0.267}{5.0} \approx 0.0534 \, \text{M} \]

### Final Answer
The equilibrium concentration of \( CO_2(g) \) is approximately \( 0.0534 \, \text{M} \).

---