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, we need to determine the amount of AB at equilibrium for the reaction:

\[ A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \]

Given:
- The equilibrium constant \( K_c = 50 \)
- Initial moles of \( A_2 = 1 \) mol
- Initial moles of \( B_2 = 2 \) mol
- Initial moles of \( AB = 0 \) mol

### Step 1: Set up the initial concentrations
Initially, we have:
- \( [A_2] = 1 \) mol
- \( [B_2] = 2 \) mol
- \( [AB] = 0 \) mol

### Step 2: Define the change in concentration
Let \( x \) be the amount of \( A_2 \) and \( B_2 \) that reacts to form \( AB \). The changes in concentration will be:
- \( A_2 \) decreases by \( x \): \( 1 - x \)
- \( B_2 \) decreases by \( x \): \( 2 - x \)
- \( AB \) increases by \( 2x \): \( 0 + 2x \)

### Step 3: Write the equilibrium expression
The equilibrium constant \( K_c \) is given by the expression:

\[ K_c = \frac{[AB]^2}{[A_2][B_2]} \]

Substituting the equilibrium concentrations into the equation:

\[ 50 = \frac{(2x)^2}{(1 - x)(2 - x)} \]

### Step 4: Simplify the equation
This simplifies to:

\[ 50 = \frac{4x^2}{(1 - x)(2 - x)} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives:

\[ 50(1 - x)(2 - x) = 4x^2 \]

Expanding the left side:

\[ 50(2 - 3x + x^2) = 4x^2 \]

This simplifies to:

\[ 100 - 150x + 50x^2 = 4x^2 \]

### Step 6: Rearrange the equation
Rearranging gives:

\[ 50x^2 - 4x^2 - 150x + 100 = 0 \]

This simplifies to:

\[ 46x^2 - 150x + 100 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 46 \), \( b = -150 \), and \( c = 100 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-150)^2 - 4 \cdot 46 \cdot 100 = 22500 - 18400 = 4100 \]

Now substituting into the quadratic formula:

\[ x = \frac{150 \pm \sqrt{4100}}{2 \cdot 46} \]

Calculating \( \sqrt{4100} \approx 64.031 \):

\[ x = \frac{150 \pm 64.031}{92} \]

Calculating the two possible values for \( x \):

1. \( x = \frac{214.031}{92} \approx 2.33 \) (not possible since it exceeds initial moles)
2. \( x = \frac{85.969}{92} \approx 0.93 \)

### Step 8: Calculate the amount of AB at equilibrium
The amount of \( AB \) formed is:

\[ [AB] = 2x = 2 \cdot 0.93 = 1.86 \text{ mol} \]

### Final Answer
The amount of \( AB \) at equilibrium is approximately **1.86 mol**.

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