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Question
Find out the value of Kc for each of the following equilibria from the value of Kp:
\[\ce{CaCO3 (s) ⇌ CaO(s) + CO2(g)}\]; Kp= 167 at 1073 K
Solution
Here,
Δn = 2 - 1 = 1
R = 0.0831 barLmol-1K-1
T = 1073 K
Kp= 167
Now,
Kp = Kc (RT) Δn
`=> 167 = "K"_"c"(0.0831 xx 1073)^(triangle"n")`
`=> "K"_"c" = 167/(0.0831 xx 1073)`
= 1.87 (approximately)
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