Kc = 0.0198 at 721K for the rxn: 2HI(g) <=> H2(g) + I2(g) In an experiment, the
ID: 75424 • Letter: K
Question
Kc = 0.0198 at 721K for the rxn: 2HI(g) <=> H2(g) + I2(g) In an experiment, the partial presssures of H2 and I2 atequilibrium are 0.710 and 0.888 atm respectively. The partialpressure of HI is ___________ atm. (a) 1.98 (b) 0.125 (c) 7.87 (c) 0.389 (d) 5.64 Please note: must Kp be converted into Kc? Kc = 0.0198 at 721K for the rxn: 2HI(g) <=> H2(g) + I2(g) In an experiment, the partial presssures of H2 and I2 atequilibrium are 0.710 and 0.888 atm respectively. The partialpressure of HI is ___________ atm. (a) 1.98 (b) 0.125 (c) 7.87 (c) 0.389 (d) 5.64 Please note: must Kp be converted into Kc?Explanation / Answer
We Know that : The given Reactionis : 2HI(g) <=> H2(g) + I2(g) Kp = Kc (RT) ng ng = 2-2 = 0 Kp = Kc Kp = PH2 PI2 / P2HI 0.0198 = 0.710 x 0.888 / P2HI PHI = 5.6429 atm option d is correct. PHI = 5.6429 atm option d is correct.Related Questions
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