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Calculate the equilibrium constant (Kc) of the reaction below at 25°C with and without a catalyst and determine the effect of the catalyst on the equilibrium position.N2(g) + 3H2(g) ⇄ 2NH3(g)

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To calculate the equilibrium constant  Kc  of the reaction at 25C, we need the standard Gibbs free energy change  G  of the reaction. For the given reaction:N2 g  + 3H2 g   2NH3 g The standard Gibbs free energy change  G  at 25C is -33.6 kJ/mol. Now, we can use the following equation to calculate the equilibrium constant  Kc :G = -RT ln Kc Where:G = -33.6 kJ/molR = 8.314 J/ molK   gas constant T = 25C = 298.15 KFirst, we need to convert G to J/mol:G = -33.6 kJ/mol * 1000 J/kJ = -33600 J/molNow, we can rearrange the equation to solve for Kc:ln Kc  = -G /  RT Kc = e^-G /  RT  Plugging in the values:Kc = e^- -33600 J/mol  /  8.314 J/ molK  * 298.15 K  Kc  6.02 x 10^2So, the equilibrium constant  Kc  for the reaction at 25C is approximately 6.02 x 10^2.Now, let's consider the effect of a catalyst on the equilibrium position. A catalyst speeds up the rate of a reaction by lowering the activation energy but does not affect the equilibrium constant  Kc . Therefore, the equilibrium constant  Kc  will remain the same, 6.02 x 10^2, with or without a catalyst.In conclusion, the equilibrium constant  Kc  for the reaction at 25C is approximately 6.02 x 10^2, and the presence of a catalyst does not affect the equilibrium position.

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