Question

The equation for the formation of hydrogen iodide from H₂ and I₂ is:

H₂(g) + I₂(g) <--> 2HI(g)

The value of K_p for the reaction is 69.0 at 730.0C. What is the equilibrium partial pressure of HI in a sealed reaction vessel at 730.0C if the initial partial pressures of H₂ and I₂ are both 0.1600 atm and initially there is no HI present?

Answer 1

write out the Kp expression, where Pp is the partial pressure:

Kp = (Pp HI^2)/(Pp H2 * Pp I2)

Pp H2 = Pp I2 =0.016 atm

Let the partial pressure of HI at equilibrium is x

So

69.0 = (x)^2/(0.16 * 0.16)

=> x² = 69 / 0.16*0.16

=> x = 1.329 atm

. To check, plug the equilibrium pressure back into the Kp expression, making sure that you get close to the Kp value you were given. If you do, you know it's correct.

Kp = 68.993 by putting the va;ue back in the equation