At 400 K, an equilibrium mixture of H 2 , I 2 , and HI consists of 0.015 mol H 2

Question

At 400 K, an equilibrium mixture of H2, I2, and HI consists of 0.015 mol H2, 0.012 mol I2, and 0.025 mol HI in a 4.50-L flask. What is the value of Kp for the following equilibrium? (R = 0.0821 L At 400 K, an equilibrium mixture of H2, I2, and HI consists of 0.015 mol H2, 0.012 mol I2, and 0.025 mol HI in a 4.50-L flask. What is the value of Kp for the following equilibrium? (R = 0.0821 L At 400 K, an equilibrium mixture of H2, I2, and HI consists of 0.015 mol H2, 0.012 mol I2, and 0.025 mol HI in a 4.50-L flask. What is the value of Kp for the following equilibrium? (R = 0.0821 L At 400 K, an equilibrium mixture of H2, I2, and HI consists of 0.015 mol H2, 0.012 mol I2, and 0.025 mol HI in a 4.50-L flask. What is the value of Kp for the following equilibrium? (R = 0.0821 L · atm/(K · mol))2HI(g) H2(g) + I2(g) 3.4 44 0.29 0.0072 138

Explanation / Answer

Kp =Kc*(RT)^n

n =2-2 =0

hence Kp =Kc

Kp=Kc= (0.015)*0.012/0.025^2 =0.29

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