The equilibrium constant at `298K` for a reaction, `A+BhArrC+D` is 100. If the initial concentrations of all the four species were 1M each, then equilibirum concentration of `D` (in mol`L^(-1)`) will be

Options

A. 0.818
B. 1.818
C. 1.182
D. 0.182

Answer

To find the equilibrium concentration of D in the reaction \( A + B \rightleftharpoons C + D \) with an equilibrium constant \( K = 100 \) at \( 298K \), we can follow these steps:

### Step 1: Write the expression for the equilibrium constant
The equilibrium constant \( K \) for the reaction can be expressed as:
\[ K = \frac{[C][D]}{[A][B]} \]
Given that \( K = 100 \).

### Step 2: Set up initial concentrations
At the start (initially, at \( t = 0 \)):
\[ [A] = 1 \, \text{M}, \quad [B] = 1 \, \text{M}, \quad [C] = 1 \, \text{M}, \quad [D] = 1 \, \text{M} \]

### Step 3: Define the change in concentrations at equilibrium
Let \( x \) be the change in concentration of A and B that reacts to form C and D. Therefore, at equilibrium:
\[ [A] = 1 - x, \quad [B] = 1 - x, \quad [C] = 1 + x, \quad [D] = 1 + x \]

### Step 4: Substitute equilibrium concentrations into the equilibrium expression
Substituting the equilibrium concentrations into the expression for \( K \):
\[ 100 = \frac{(1 + x)(1 + x)}{(1 - x)(1 - x)} \]
This simplifies to:
\[ 100 = \frac{(1 + x)^2}{(1 - x)^2} \]

### Step 5: Take the square root of both sides
Taking the square root of both sides gives:
\[ 10 = \frac{1 + x}{1 - x} \]

### Step 6: Solve for \( x \)
Cross-multiplying gives:
\[ 10(1 - x) = 1 + x \]
Expanding and rearranging:
\[ 10 - 10x = 1 + x \\ 10 - 1 = 10x + x \\ 9 = 11x \\ x = \frac{9}{11} \approx 0.818 \]

### Step 7: Find the equilibrium concentration of D
Now, substituting \( x \) back to find the concentration of D:
\[ [D] = 1 + x = 1 + 0.818 = 1.818 \, \text{M} \]

### Final Answer
The equilibrium concentration of D is:
\[ \boxed{1.818 \, \text{mol L}^{-1}} \]