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Ch.15 - Chemical Equilibrium

Chapter 15, Problem 62a

The reaction 2 AsH31g2 ∆ As21g2 + 3 H21g2 has Kp = 7.2 * 107 at 1073 K. At the same temperature, what is Kp for each of the following reactions? (a) As21g2 + 3 H21g2 ∆ 2 AsH31g2

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Hello. In this problem, we are told that it's 600 kelvin. The reaction between hydrogen iodine form hydrogen iodide has an equilibrium constant of 35.5. Were asked. What is the value of the equilibrium constant for the reaction for the decomposition of hydrogen iodide to form hydrogen iodine at the same temperature. So that's right. That equilibrium constant expression for our first reaction. So K P then is equal to the partial pressure of hydrogen iodide and that's squared based on the cold fish. And the balanced reaction equation divided by the partial pressure of hydrogen. The one power times the partial pressure of iodine to the one power We're told this is equal to 35.5. Now that's right, the equilibrium constant expression will call that K. P. Prime for our second reaction. So we have the partial pressure of our products. So hydrogen and iodine the one power based on the coefficient in the balanced reaction equation divided by the partial pressure hydrogen iodide squared. So when we compare K P two K P prime, we can see that they are the inverse of each other. So K. P prime is equal to one over K P. So that means then the equilibrium constant value for our second reaction will be one over that for the first One over 35.5, this works out to 0.0282. So our equilibrium constant for a reaction written in reverse is equal to one over the equilibrium constant for the reaction in the forward direction. Thanks for watching. Hope this helps