At a given temperature, Kc is 4 for the reaction:
`H_2(g)+CO_2(g) ⇔H_2O(g)+CO(g)`.Initially 0.6 moles each of` H_2` and `CO_2` are taken in 1 liter flask. The equilibrium concentration of `H_2O(g)` is :

Explanation

To solve the problem, we will follow these steps:

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

### Step 2: Set up the initial concentrations.
Initially, we have:
- \([H_2] = 0.6 \, \text{mol/L}\)
- \([CO_2] = 0.6 \, \text{mol/L}\)
- \([H_2O] = 0 \, \text{mol/L}\)
- \([CO] = 0 \, \text{mol/L}\)

### Step 3: Define the change in concentration at equilibrium.
Let \(x\) be the change in concentration of \(H_2O\) and \(CO\) formed at equilibrium. Thus, we can express the equilibrium concentrations as:
- \([H_2] = 0.6 - x\)
- \([CO_2] = 0.6 - x\)
- \([H_2O] = x\)
- \([CO] = x\)

### Step 4: Write the expression for the equilibrium constant \(K_c\).
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[H_2O][CO]}{[H_2][CO_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{x \cdot x}{(0.6 - x)(0.6 - x)} = \frac{x^2}{(0.6 - x)^2} \]

### Step 5: Substitute the value of \(K_c\).
Given that \(K_c = 4\), we can set up the equation:
\[ 4 = \frac{x^2}{(0.6 - x)^2} \]

### Step 6: Cross-multiply and simplify.
Cross-multiplying gives:
\[ 4(0.6 - x)^2 = x^2 \]
Expanding the left side:
\[ 4(0.36 - 1.2x + x^2) = x^2 \]
\[ 1.44 - 4.8x + 4x^2 = x^2 \]

### Step 7: Rearrange the equation.
Rearranging gives:
\[ 4x^2 - x^2 - 4.8x + 1.44 = 0 \]
\[ 3x^2 - 4.8x + 1.44 = 0 \]

### Step 8: Solve the quadratic equation.
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 3\), \(b = -4.8\), and \(c = 1.44\):
\[ x = \frac{-(-4.8) \pm \sqrt{(-4.8)^2 - 4 \cdot 3 \cdot 1.44}}{2 \cdot 3} \]
Calculating the discriminant:
\[ (-4.8)^2 - 4 \cdot 3 \cdot 1.44 = 23.04 - 17.28 = 5.76 \]
Now substituting back:
\[ x = \frac{4.8 \pm \sqrt{5.76}}{6} \]
Calculating \(\sqrt{5.76} = 2.4\):
\[ x = \frac{4.8 \pm 2.4}{6} \]
Calculating the two possible values for \(x\):
1. \(x = \frac{7.2}{6} = 1.2\) (not possible since initial concentration is 0.6)
2. \(x = \frac{2.4}{6} = 0.4\)

### Step 9: Determine the equilibrium concentration of \(H_2O\).
Thus, the equilibrium concentration of \(H_2O\) is:
\[ [H_2O] = x = 0.4 \, \text{mol/L} \]

### Final Answer:
The equilibrium concentration of \(H_2O(g)\) is \(0.4 \, \text{mol/L}\).

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