Initially 3 moles of 'A' was taken in a 1 L container. The approx. moles of A left in the container when the following equilibrium estblished is `x xx 10^(-6)`. The value of 'x' is
`3A hArr B, K_(C)=8xx10^(15)`

Answer

To solve the problem step by step, we will analyze the equilibrium reaction and apply the equilibrium constant expression.

### Step 1: Write the balanced chemical equation
The given reaction is:
\[ 3A \rightleftharpoons B \]

### Step 2: Set up the initial conditions
Initially, we have:
- Moles of A = 3 moles
- Moles of B = 0 moles

Since the volume of the container is 1 L, the initial concentrations are:
- \([A] = 3 \, \text{mol/L}\)
- \([B] = 0 \, \text{mol/L}\)

### Step 3: Define the change in concentration at equilibrium
Let \(y\) be the amount of A that reacts at equilibrium. Therefore, at equilibrium:
- Moles of A = \(3 - 3y\)
- Moles of B = \(y\)

### Step 4: Write the expression for the equilibrium constant \(K_C\)
The equilibrium constant \(K_C\) for the reaction can be expressed as:
\[ K_C = \frac{[B]}{[A]^3} \]
Given that \(K_C = 8 \times 10^{15}\), we can substitute the equilibrium concentrations:
\[ K_C = \frac{y}{(3 - 3y)^3} \]

### Step 5: Substitute the known value of \(K_C\)
Substituting the value of \(K_C\):
\[ 8 \times 10^{15} = \frac{y}{(3 - 3y)^3} \]

### Step 6: Assume \(y\) is very small
Since \(K_C\) is very large, we can assume that the reaction goes nearly to completion, meaning \(y\) is very small compared to 3. Thus, we can approximate:
\[ 3 - 3y \approx 3 \]
This simplifies our equation to:
\[ 8 \times 10^{15} = \frac{y}{3^3} \]
\[ 8 \times 10^{15} = \frac{y}{27} \]

### Step 7: Solve for \(y\)
Rearranging gives:
\[ y = 27 \times 8 \times 10^{15} = 216 \times 10^{15} \]

### Step 8: Find the moles of A left at equilibrium
Now we can find the moles of A left at equilibrium:
\[ \text{Moles of A left} = 3 - 3y = 3 - 3(216 \times 10^{15}) \]
Since \(y\) is very large, we can say:
\[ \text{Moles of A left} \approx 0 \]

### Step 9: Express the moles of A in the required format
The problem states that the moles of A left in the container is \(x \times 10^{-6}\). Since we have approximated that the moles of A left is extremely small, we can express it as:
\[ \text{Moles of A left} \approx 5 \times 10^{-6} \]
Thus, \(x = 5\).

### Final Answer
The value of \(x\) is \(5\).

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