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 \( K_c = 100 \) at \( 298 \, K \), and initial concentrations of all species at \( 1 \, M \), we can follow these steps:

### Step 1: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) for the reaction can be expressed as:
\[ K_c = \frac{[C][D]}{[A][B]} \]
Given that \( K_c = 100 \), we can substitute this into our equation:
\[ 100 = \frac{[C][D]}{[A][B]} \]

### Step 2: Set up the initial concentrations
Initially, we have:
\[ [A]_0 = 1 \, M, \quad [B]_0 = 1 \, M, \quad [C]_0 = 1 \, M, \quad [D]_0 = 1 \, M \]

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

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

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 100(1 - x)^2 = (1 + x)^2 \]
Expanding both sides:
\[ 100(1 - 2x + x^2) = 1 + 2x + x^2 \]
This simplifies to:
\[ 100 - 200x + 100x^2 = 1 + 2x + x^2 \]
Rearranging gives:
\[ 99x^2 - 202x + 99 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{202 \pm \sqrt{(-202)^2 - 4 \cdot 99 \cdot 99}}{2 \cdot 99} \]
Calculating the discriminant:
\[ (-202)^2 - 4 \cdot 99 \cdot 99 = 40804 - 39204 = 1600 \]
Thus,
\[ x = \frac{202 \pm 40}{198} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{242}{198} = \frac{121}{99} \approx 1.222 \)
2. \( x = \frac{162}{198} = \frac{81}{99} \approx 0.818 \)

### Step 7: Calculate the equilibrium concentration of D
Since \( D = 1 + x \), we will use the value of \( x \) that is less than 1 (as concentrations cannot exceed initial concentrations):
\[ x = \frac{81}{99} \approx 0.818 \]
Thus, the equilibrium concentration of D is:
\[ [D] = 1 + x = 1 + \frac{81}{99} = \frac{99 + 81}{99} = \frac{180}{99} \approx 1.818 \, M \]

### Final Answer
The equilibrium concentration of D is approximately \( 1.818 \, M \).

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