The equilibrium constant `(K_(c))` of the reaction `A_(2)(g) + B_(2)(g) rarr 2AB(g)` is 50. If 1 mol of `A_(2)` and 2 mol of `B_(2)` are mixed, the amount of AB at equilibrium would be

Options

A. 0.934 mol
B. 0.467 mol
C. 1.866 mol
D. 1.401 mol

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Write the balanced chemical equation and identify initial concentrations.
The balanced chemical equation for the reaction is:
\[ A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \]

Initially, we have:
- Moles of \( A_2 = 1 \) mol
- Moles of \( B_2 = 2 \) mol
- Moles of \( AB = 0 \) mol

### Step 2: Set up the ICE table (Initial, Change, Equilibrium).
We can set up an ICE table to track the changes in concentration:

| Species | Initial (mol) | Change (mol) | Equilibrium (mol) |
|---------|----------------|---------------|--------------------|
| \( A_2 \) | 1 | -x | \( 1 - x \) |
| \( B_2 \) | 2 | -x | \( 2 - x \) |
| \( AB \) | 0 | +2x | \( 2x \) |

### Step 3: Write the expression for the equilibrium constant \( K_c \).
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[AB]^2}{[A_2][B_2]} \]
Substituting the equilibrium concentrations from the ICE table:
\[ K_c = \frac{(2x)^2}{(1 - x)(2 - x)} \]

### Step 4: Substitute the value of \( K_c \) and solve for \( x \).
Given that \( K_c = 50 \), we can set up the equation:
\[ 50 = \frac{4x^2}{(1 - x)(2 - x)} \]

### Step 5: Cross-multiply and simplify the equation.
Cross-multiplying gives:
\[ 50(1 - x)(2 - x) = 4x^2 \]
Expanding the left-hand side:
\[ 50(2 - 3x + x^2) = 4x^2 \]
This simplifies to:
\[ 100 - 150x + 50x^2 = 4x^2 \]
Rearranging gives:
\[ 46x^2 - 150x + 100 = 0 \]

### Step 6: Solve the quadratic equation.
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 46, b = -150, c = 100 \):
\[ x = \frac{150 \pm \sqrt{(-150)^2 - 4 \cdot 46 \cdot 100}}{2 \cdot 46} \]
Calculating the discriminant:
\[ x = \frac{150 \pm \sqrt{22500 - 18400}}{92} \]
\[ x = \frac{150 \pm \sqrt{4100}}{92} \]
Calculating \( \sqrt{4100} \approx 64.03 \):
\[ x = \frac{150 \pm 64.03}{92} \]
Calculating the two possible values for \( x \):
1. \( x \approx \frac{214.03}{92} \approx 2.32 \) (not valid since it exceeds initial moles)
2. \( x \approx \frac{85.97}{92} \approx 0.934 \)

### Step 7: Calculate the amount of \( AB \) at equilibrium.
The amount of \( AB \) at equilibrium is given by:
\[ \text{Moles of } AB = 2x = 2 \times 0.934 \approx 1.868 \text{ moles} \]

### Final Answer:
The amount of \( AB \) at equilibrium is approximately **1.868 moles**.

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