$$PCl_5 \rightleftharpoons PCl_3 + Cl_2 \quad K_c = 1.844$$
3.0 moles of PCl$$_5$$ is introduced in a 1L closed reaction vessel at 380 K. The number of moles of PCl$$_5$$ at equilibrium is _________ $$\times 10^{-3}$$ (Round off to the Nearest Integer)

We consider the equilibrium reaction

$$PCl_5\; \rightleftharpoons\; PCl_3 + Cl_2$$

with the given equilibrium constant

$$K_c = 1.844$$

Initially, 3.0 moles of $$PCl_5$$ are placed in a 1 L vessel at 380 K. Because the volume is 1 L, the numerical value of the concentration equals the number of moles. Hence the initial concentrations are

$$[PCl_5]_0 = 3.0\;{\rm M}, \qquad [PCl_3]_0 = 0,\qquad [Cl_2]_0 = 0.$$

Let $$x$$ be the number of moles of $$PCl_5$$ that dissociate at equilibrium. The changes and the equilibrium amounts are therefore

$$\begin{aligned} PCl_5 &: \; 3.0 - x,\\ PCl_3 &: \; 0 + x = x,\\ Cl_2 &: \; 0 + x = x. \end{aligned}$$

Again, because the volume is 1 L, these numbers are simultaneously the equilibrium concentrations.

The law of mass action states that for the reaction $$PCl_5 \rightleftharpoons PCl_3 + Cl_2$$ the equilibrium constant is

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

Substituting the equilibrium concentrations we obtain

$$1.844 = \frac{(x)(x)}{3.0 - x} = \frac{x^2}{3.0 - x}.$$

Cross-multiplying gives

$$x^2 = 1.844\,(3.0 - x).$$

Expanding the right-hand side,

$$x^2 = 1.844 \times 3.0 - 1.844\,x = 5.532 - 1.844\,x.$$

Bringing every term to the left produces a quadratic equation,

$$x^2 + 1.844\,x - 5.532 = 0.$$

We now solve this quadratic using the quadratic formula. For an equation $$ax^2 + bx + c = 0,$$ the roots are

$$x = \frac{-b \pm \sqrt{\,b^2 - 4ac\,}}{2a}.$$

Here $$a = 1,\; b = 1.844,\; c = -5.532.$$ Substituting these values,

$$\begin{aligned} x &= \frac{-1.844 \pm \sqrt{(1.844)^2 - 4(1)(-5.532)}}{2(1)}\\[4pt] &= \frac{-1.844 \pm \sqrt{3.400336 + 22.128}}{2}\\[4pt] &= \frac{-1.844 \pm \sqrt{25.528336}}{2}\\[4pt] &= \frac{-1.844 \pm 5.052}{2}. \end{aligned}$$

The negative sign would give a negative value of $$x$$, which is physically impossible, so we take the positive sign:

$$x = \frac{-1.844 + 5.052}{2} = \frac{3.208}{2} = 1.604.$$

Therefore, the equilibrium amount of $$PCl_5$$ is

$$3.0 - x = 3.0 - 1.604 = 1.396\;{\rm moles}.$$

Expressing this in the form “number × 10−3”,

$$1.396\;{\rm moles} = 1.396 \times 10^{0}\;{\rm moles} = 1396 \times 10^{-3}\;{\rm moles}.$$

Rounding 1396 to the nearest integer yields 1400.

Hence, the correct answer is Option 1400.