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A reaction mixture at equilibrium at 175 k contains ph2 = 0.958 atm, pi2 = 0.877 atm, and phi = 0.020 atm. a second reaction mixture, also at 175 k, contains ph2 = pi2 = 0.621 atm, and phi = 0.101 atm. is the second reaction at equilibrium? if not, what will be the partial pressure of hi when the reaction reaches equilibrium at 175 k?

Dana Keller

in Chemistry

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Jeffrey Rodriguez on August 11, 2018

ICE table H2(g) + I2 (g )↔ 2HI(g) equ 0.958 0.877 0.02 first mix1 0.621 0.621 0.101 sec mix2Kp1 = P(HI)^2 / p(A2)*p(I2) to the blend 1 = 0.02^2 / 0.958*0.877 = 4.8x10^-4Kp2 = P(HI)^2 / P(A2)* P(I2) to the blend 2 = 0.101^2/ 0.621*0.621 = 0.0265 we can see that Kp1< Kp2 which means that the sec mixture is not in balance. Going to go to the left to reduce its products and to increase the reactant to reduce the value of Kp to achieve the balance.and the partial pressure of Hi when the mixture 2 to reach the equilibrium is:4.8x10^-4 = P(Hi)^2 / (0.621*0.621)∴ P(Hi) at equilibrium = 0.0136 atm


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