Question

$PCl_5 \rightleftharpoons PCl_3 + Cl_2 K_c = 1.844$
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 _________ $\times 10^{-3}$. (Round off to the Nearest Integer)

Hint:

For equilibrium problems in a 1 L vessel, molarity equals number of moles. Always discard the negative root of a quadratic equation, as concentration cannot be negative.

Answer 1

Correct Answer: 1258

The equilibrium reaction is:
\[ \mathrm{PCl_5 \rightleftharpoons PCl_3 + Cl_2} \] Given: \[ K_c = 1.844 \] Volume of vessel = 1 L 
Initial moles of $\mathrm{PCl_5}$ = 3.0 
Since volume is 1 L, molarity equals number of moles. 
Step 1: ICE table 

Step 2: Write equilibrium constant expression \[ K_c = \frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]} \] \[ 1.844 = \frac{x^2}{3-x} \] Step 3: Solve the equation \[ 1.844(3-x) = x^2 \] \[ x^2 + 1.844x - 5.532 = 0 \] Solving using quadratic formula: \[ x = \frac{-1.844 + \sqrt{(1.844)^2 + 4(5.532)}}{2} \] \[ x \approx 1.74 \] Step 4: Moles of $\mathrm{PCl_5}$ at equilibrium \[ n_{\mathrm{PCl_5}} = 3.0 - x \] \[ = 3.0 - 1.74 = 1.26 \text{ mol} \] Step 5: Express in required format \[ 1.26 \text{ mol} = 1258 \times 10^{-3} \text{ mol (rounded)} \] \[ \boxed{1258} \]