At 400°C the reaction H2+I2<->2HI has an equilibrium constant Kp value of 55.5. A mixture of 1 mole of H2 and 1 mole of I2 is placed in a flask and heated to 400°C. Calculate the mole fraction of HI in the mixture of gases when equilibrium reached.

Answer 1

55.3=(2x)^2/(1-x)^2

solve for x Use the quadratic equation

Answer 2

i don't understand bobpursley..how is it actually?

Answer 3

H2 +I2---> 2HI.

The equilibrium expression is

Kp = reactants/products=[2HI]^2/[H2][I2]

Since you have 2 moles of HI for every mole of H2 or I2, you square the products.

You start off with 1 mole of H2 and I2, and some, not all of the reaction produces x amount of HI. You don't know how much, but you know the molar ratio of HI will be twice as much as H2 or I2, so that is why he told you (2x)^2 for the product. Since the amount of HI is subtracted from the initial concentrations of H2 and I2 to make HI, the amount subtracted is 1-X for each. Looking at the equation that I gave you above, you will see that reactants are multiplied by each other to produce (1-x)^2. The rest is just using algebra to solve for x, and the only way that you can do that is to use the quadratic equation.

I hope this helps.

Answer 4

Its helps a lot...btw are you assuming Kp=Kc?

Answer 5

Yes.

Answer 6

Sorry for asking such simple question. Why we must assume like that? And why not the Kp= [HI]^2/[H2][I2]?Yet you put 2 in front of HI. Typo?