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09/10/2021 11:21 am
Topic starter
At 527°C, the reaction given below has Kc = 4
NH3(g) ⇌ 1/2N2(g) + 3/2H2(g)
What is the Kp for the reaction?
N2(g) + 3H2(g) ⇌ 2NH3(g)
(a) 16 x (800R)2
(b) \(\Big(\frac{800R}{4}\Big)^{-2}\)
(c) \(\Big(\frac{1}{4 \times 800R}\Big)^{2}\)
(d) None of these
1 Answer
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09/10/2021 11:28 am
Correct answer: (c) \(\Big(\frac{1}{4 \times 800R}\Big)^{2}\)
Explanation:
If, Kc = \(\frac{[N_2]^{\frac{1}{2}}[H_2]^{\frac{3}{2}}}{[NH_3]^1}\) = 4
then, \(\frac{[NH_3]^2}{[N_2][H_2]^3} = (4)^{-2}\)
= \(\frac{1}{16}\)
also Kp = Kc .(RT)Δng
= \(\frac{1}{16} \times (800R)^{-2}\)
= \(\Big(\frac{1}{4 \times 800R}\Big)^{2}\)