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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry, 4th edition.

Problem 14.20.

Determine the values of Kp that correspond to the following values of Kc.

(a) CO(g) + Cl2(g)COCl2(g)

Kc = 1.2×103 at 668 K

(b) 2 NO(g) + Br2(g)2 NOBr(g)

Kc = 1.32×10–2 at 1000 K

(c) 2 COF2(g)CO2(g) + CF4(g)

Kc = 2.00 at 1000 oC




Answer:

Use the relationship Kp = Kc(RT)n

(a) n = 1 – 2 = –1, so Kp = 1.2×103(0.0821×668)–1 = 22

(b) n = 2 – 3 = –1, so Kp = 1.32×10–2(0.0821×1000)–1 = 1.61×10–4

(c) n = 2 – 2 = 0, T = 1000 + 273 = 1273 K so Kp = 2.00(0.0821×1273)0 = 2.00