Question

PCl3(g)+Cl2(g)<---->PCl5(g) Kc=20 @ 240 C

If a 1.0 L container contains 0.25 mol of PCl5 at 240 C, how many moles of it dissociate to form PCl3 and Cl2?

Answer 1

Answer on Question#51004 – Chemistry – Other

$$\mathrm{PCl3(g)} + \mathrm{Cl2(g)} < \dots < \mathrm{gt}; \mathrm{PCl5(g)} \quad \mathrm{Kc} = 20 @ 240^\circ \mathrm{C}$$

If a 1.0 L container contains 0.25 mol of PCl5 at 240 C, how many moles of it dissociate to form PCl3 and Cl2?

**Solution:**

$$aA = bB + cC; \quad Kc = [B]^b [C]^c / [A]^a;$$ $$Kc \text{ – the equilibrium constant (the concentration quotient);}$$ $$[ ] \text{ – the molar concentration (mol/L);}$$ $$C = n/V;$$ $$C \text{ – molar concentration (mol/L); } n \text{ – mole (mol); } V \text{ – volume (L);}$$ $$\text{Reaction: } \mathrm{PCl5(g)} = \mathrm{PCl3(g)} + \mathrm{Cl2(g)};$$

According to the equilibrium: $n(\mathrm{PCl5}) : n(\mathrm{PCl3}) : n(\mathrm{Cl2}) = 1 : 1 : 1$;

$$C(\mathrm{PCl5}) = n(\mathrm{PCl5})/V; \quad C(\mathrm{PCl5}) = 0.25 \text{ mol/L};$$

Initial concentration $\mathrm{PCl5} = 0.25\ \mathrm{M}; \quad Kc = 20$;

$$x = n(\mathrm{PCl5}) \text{ – that dissociated; } (0.25 - x) = n \text{ (equilibrium);}$$ $$V \text{ – constant; } n(\mathrm{PCl3}) = n(\mathrm{Cl2}) = n(\mathrm{PCl5})_{\mathrm{dis}} = x;$$ $$Kc = \frac{C(\mathrm{PCl3}) \cdot C(\mathrm{Cl2})}{C(\mathrm{PCl5})_{eq}} = \frac{(n(\mathrm{PCl3})/V) \cdot (n(\mathrm{Cl2})/V)}{n(\mathrm{PCl5})_{eq}/V} = \frac{n(\mathrm{PCl3}) \cdot n(\mathrm{Cl2})}{n(\mathrm{PCl5})_{eq}};$$ $$Kc = x^2 / (0.25 - x);$$ $$x^2 / (0.25 - x) = 20;$$ $$x^2 + 20x - 5 = 0;$$ $$x = 0.25;$$ $$n(\mathrm{PCl5})_{\mathrm{dis}} = 0.25 \text{ mol}$$

Answer: 0.25 mol.

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