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the equilibrium constant kp for the following reaction at 191 degree celcius is 1.24. what is kc ? b(s) + 3/2f 2(g) = BF 3 the equilibrium constant kp for the following reaction at 191 degree celcius is 1.24. what is kc ? b(s) + 3/2f2(g) = BF3
Dear Rahul Use formula Kp = Kc * (RT)^(delta n) Now I hope you can do it. In case of any difficulty, Please feel free to ask.
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