`PCl_(5) hArr PCl_(3) + Cl_(3) " "K_(c)= 1.844`
3.0 moles of `PCl_(5)` is introduced in a 1 L closed reaction vessel at 380 K. The number of moles of `PCl_(5)` at equilibrium is ________ ` xx 10^(-3)`.
(Round off to the Nearest Integer)

Answer

To solve the problem, we need to determine the number of moles of \( PCl_5 \) at equilibrium in the reaction:

\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

Given:
- Initial moles of \( PCl_5 = 3.0 \) moles
- \( K_c = 1.844 \)
- Volume of the reaction vessel = 1 L

### Step 1: Set up the initial conditions
Initially, we have:
- \( [PCl_5] = 3.0 \) M (since the volume is 1 L)
- \( [PCl_3] = 0 \) M
- \( [Cl_2] = 0 \) M

### Step 2: Define the change in concentration
Let \( x \) be the number of moles of \( PCl_5 \) that dissociate at equilibrium. Therefore, at equilibrium, we have:
- \( [PCl_5] = 3.0 - x \)
- \( [PCl_3] = x \)
- \( [Cl_2] = x \)

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

\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]

Substituting the equilibrium concentrations into the expression, we get:

\[ 1.844 = \frac{x \cdot x}{3.0 - x} = \frac{x^2}{3.0 - x} \]

### Step 4: Rearrange the equation
Multiplying both sides by \( (3.0 - x) \):

\[ 1.844(3.0 - x) = x^2 \]

Expanding this gives:

\[ 5.532 - 1.844x = x^2 \]

Rearranging it into standard quadratic form:

\[ x^2 + 1.844x - 5.532 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1.844, c = -5.532 \):

1. Calculate the discriminant \( D \):

\[ D = b^2 - 4ac = (1.844)^2 - 4 \cdot 1 \cdot (-5.532) \]
\[ D = 3.396736 + 22.128 = 25.524736 \]

2. Now calculate \( x \):

\[ x = \frac{-1.844 \pm \sqrt{25.524736}}{2} \]
\[ x = \frac{-1.844 \pm 5.052}{2} \]

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

- \( x_1 = \frac{3.208}{2} = 1.604 \)
- \( x_2 = \frac{-6.896}{2} \) (not physically meaningful since concentration cannot be negative)

Thus, \( x \approx 1.604 \).

### Step 6: Calculate moles of \( PCl_5 \) at equilibrium
At equilibrium, the moles of \( PCl_5 \) are:

\[ [PCl_5] = 3.0 - x = 3.0 - 1.604 = 1.396 \]

### Step 7: Express the answer in the required format
To express this in the format \( xx \times 10^{-3} \):

\[ 1.396 \text{ moles} = 1396 \times 10^{-3} \text{ moles} \]

Rounding off to the nearest integer gives:

\[ \text{Final answer} = 1396 \]