The equilibrium constant Kc for the reaction shown below is 3.79 x 10^-5 at 727

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The equilibrium constant Kc for the reaction shown below is 3.8 10-5 at 727°C. I2(g) ---> 2 I(g) Calculate Kc and Kp for the following equilibrium at the same temperature. 2 I(g) ---> I2(g) ***Kc for the first reaction is [I]^2/[I2] and equals 3.8 x 10^-5 Kc for the second reaction is [I2] /[I]^2, which is the inverse of the first reaction. Therefore the numerical value is also the inverse of the first, 1/(3.8 x 10^-5) = 2.6 x 10^4 Kp = Kc * (RT)^?n, where ? n = (number moles of gaseous product - number moles of gaseous reactant). In this case, there is 1 mole of gas in the product and two moles of gas in the reactants, thus ? n = -1 Kp = Kc * (RT)^? n = (2.6 x 10^4)*((0.0821)*(727 + 273.15K))^-1 = 3.2 x 10^2

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