In a container `H_(2)O(g),CO(g) "and" H_(2)(g)` are present in the molar ration of `1:2:3` respectively at temperature of `300K`, Find the pressure in the container at which solid carbon (graphile) will star forming in the container given that:
`C(S)+H_(2)O(G)hArrCO(g)+H_(2)(g) K_(p)=3` atm

Explanation

To solve the problem step-by-step, we will follow the given information and perform the necessary calculations.

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

### Step 2: Identify the molar ratios
The molar ratios of the components in the container are given as:
- \( H_2O(g) : CO(g) : H_2(g) = 1 : 2 : 3 \)

Let’s denote the partial pressures of these gases as follows:
- \( P_{H_2O} = P \)
- \( P_{CO} = 2P \)
- \( P_{H_2} = 3P \)

### Step 3: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{CO} \cdot P_{H_2}}{P_{H_2O}} \]
Substituting the expressions for the partial pressures:
\[ K_p = \frac{(2P)(3P)}{P} = \frac{6P^2}{P} = 6P \]

### Step 4: Set \( K_p \) equal to the given value
We know that \( K_p = 3 \) atm, so we set up the equation:
\[ 6P = 3 \]

### Step 5: Solve for \( P \)
To find \( P \):
\[ P = \frac{3}{6} = \frac{1}{2} \text{ atm} \]

### Step 6: Calculate the total pressure in the container
The total pressure \( P_{total} \) in the container is the sum of the partial pressures:
\[ P_{total} = P_{H_2O} + P_{CO} + P_{H_2} = P + 2P + 3P = 6P \]
Substituting the value of \( P \):
\[ P_{total} = 6 \times \frac{1}{2} = 3 \text{ atm} \]

### Conclusion
The total pressure in the container at which solid carbon (graphite) will start forming is:
\[ \boxed{3 \text{ atm}} \]