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08/10/2021 1:32 pm
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Kc for PCl5(g) ⇌ PCl3(g) + Cl2(g) is 0.04 at 250°C. How many moles of PCl5 must be added to a 3L flask to obtain a Cl2 concentration of 0.15 M
(a) 4.2 moles
(b) 2.1 moles
(c) 5.5 moles
(d) 6.3 moles
1 Answer
1
08/10/2021 1:51 pm
Correct answer: (b) 2.1 moles
Explanation:
At equilibrium the moles of Cl2 must be
= 0.15 × 3 = 0.45
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Eqn. Conc. \(\frac{x - 0.45}{3}\) \(\frac{0.45}{3}\) \(\frac{0.45}{3}\)
Kc = \(\frac{[PCl_3][Cl_2]}{[PCl_5]}\)
∴ 0.04 = \(\frac{0.15\times 0.15}{\frac{(x-0.45)}{3}}\)
∴ x = 2.1 moles