If reaction `A+BhArrC+D`, taken place in `5` liter close vessel, the rate constant of forward reaction is nine times of rate of backward reaction.
If initially one mole of each reactant present in the container, then find the correct option//is.

Options

A. `([C])/([B])=(3)/(1)`
B. `"log" K_(P)="log"K_(C)`
C. `[D]_(eq)=15xx10^(-2) "mole" L-1`
D. `K_(eq)=9`

Explanation

To solve the problem step by step, we will analyze the given reaction and the information provided.

### Step 1: Write the Reaction
The reaction given is:
\[ A + B \rightleftharpoons C + D \]

### Step 2: Initial Concentrations
Initially, we have:
- Moles of A = 1
- Moles of B = 1
- Moles of C = 0
- Moles of D = 0

Since the volume of the vessel is 5 liters, the initial concentrations are:
- \([A] = \frac{1}{5} = 0.2 \, \text{M}\)
- \([B] = \frac{1}{5} = 0.2 \, \text{M}\)
- \([C] = 0 \, \text{M}\)
- \([D] = 0 \, \text{M}\)

### Step 3: Change in Concentration at Equilibrium
Let \(x\) be the amount of A and B that reacts to form C and D at equilibrium. Therefore, at equilibrium:
- \([A] = \frac{1 - x}{5}\)
- \([B] = \frac{1 - x}{5}\)
- \([C] = \frac{x}{5}\)
- \([D] = \frac{x}{5}\)

### Step 4: Rate Constants Relation
It is given that the rate constant of the forward reaction (\(k_f\)) is nine times that of the backward reaction (\(k_b\)):
\[ k_f = 9 k_b \]

### Step 5: Equilibrium Constant Expression
The equilibrium constant (\(K_c\)) is given by:
\[ K_c = \frac{[C][D]}{[A][B]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{x}{5}\right)\left(\frac{x}{5}\right)}{\left(\frac{1-x}{5}\right)\left(\frac{1-x}{5}\right)} = \frac{x^2/25}{(1-x)^2/25} = \frac{x^2}{(1-x)^2} \]

### Step 6: Substitute \(K_c\) in Terms of \(k_f\) and \(k_b\)
From the relation of \(k_f\) and \(k_b\):
\[ K_c = \frac{k_f}{k_b} = \frac{9 k_b}{k_b} = 9 \]
Thus, we have:
\[ \frac{x^2}{(1-x)^2} = 9 \]

### Step 7: Solve for \(x\)
Taking the square root of both sides:
\[ \frac{x}{1-x} = 3 \]
Cross-multiplying gives:
\[ x = 3(1 - x) \implies x = 3 - 3x \implies 4x = 3 \implies x = \frac{3}{4} \]

### Step 8: Find Equilibrium Concentrations
Now substituting \(x\) back to find the equilibrium concentrations:
- \([A] = \frac{1 - \frac{3}{4}}{5} = \frac{1/4}{5} = \frac{1}{20} = 0.05 \, \text{M}\)
- \([B] = \frac{1 - \frac{3}{4}}{5} = 0.05 \, \text{M}\)
- \([C] = \frac{\frac{3}{4}}{5} = \frac{3/4}{5} = \frac{3}{20} = 0.15 \, \text{M}\)
- \([D] = \frac{\frac{3}{4}}{5} = 0.15 \, \text{M}\)

### Step 9: Verify Concentration Ratios
We can calculate the ratio of concentrations:
\[ \frac{[C]}{[B]} = \frac{0.15}{0.05} = 3 \]

### Step 10: Summary of Results
- \([C] = 0.15 \, \text{M}\)
- \([B] = 0.05 \, \text{M}\)
- \([D] = 0.15 \, \text{M}\)
- \(K_c = 9\)

### Conclusion
All options provided in the question are correct based on our calculations.