1. Write the equilibrium reaction and initial concentrations

$PCl_5 \rightleftharpoons PCl_3 + Cl_2$

Given:

• $K_c = 1.844$
• Volume $= 1\,L$
• Initial moles of $PCl_5 = 3.0$

Since volume is $1\,L$, initial concentration is:

$[PCl_5]_0 = 3.0\,M$

Initially:

$[PCl_5]=3.0,\quad [PCl_3]=0,\quad [Cl_2]=0$

2. Let dissociation be $x$

At equilibrium:

$[PCl_5] = 3-x$ $[PCl_3] = x$ $[Cl_2] = x$

3. Apply the equilibrium constant expression

$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{x^2}{3-x}$

Given $K_c=1.844$:

$\frac{x^2}{3-x}=1.844$

$x^2 = 1.844(3-x)$

$x^2 = 5.532 - 1.844x$

$x^2 + 1.844x - 5.532 = 0$

4. Solve the quadratic equation

$x = \frac{-1.844 \pm \sqrt{(1.844)^2 + 4(5.532)}}{2}$

First compute the discriminant:

$(1.844)^2 = 3.400336$ $4(5.532)=22.128$

So,

$D=3.400336+22.128=25.528336$

$\sqrt{D} \approx 5.05256$

Thus,

$x = \frac{-1.844 + 5.05256}{2} \approx \frac{3.20856}{2} \approx 1.60428$

(Negative root is rejected.)

5. Find equilibrium moles of $PCl_5$

$[PCl_5]_{eq} = 3 - x = 3 - 1.60428 = 1.39572$

Since volume is $1\,L$, equilibrium moles are also:

$n(PCl_5)_{eq} = 1.39572\,\text{mol}$

6. Match the required form

The question asks for the number in the form:

$\underline{\hspace{1cm}} \times 10^{-3}$

Convert moles into this form:

$1.39572 = 1395.72 \times 10^{-3}$

Rounded to nearest integer:

$1396$

7. Comparison with stored answer

Stored correct answer = $1400$

Our derived value is $1396$, which rounds from the exact calculation. The stored answer appears to be a rounded/approximated value, likely using fewer significant digits. Hence I do not exactly agree with the stored answer, though it is very close.