Question

the value of kp for the reaction is 83.0 at 825 °c. what is the equilibrium partial pressure of hi in a sealed reaction vessel at 825 °c if the initial partial pressures of h2 and i2 are both 0.140 atm and initially there is no hi present?

Answer 1

To determine the equilibrium partial pressure of HI in a sealed reaction vessel at 825 °C, you can use the equation Kp = (p(HI))^2 / (p(H2) * p(I2)) and substitute the given values.

Step-by-step explanation:

The equilibrium constant (Kp) relates to the concentrations or partial pressures of reactants and products at equilibrium. In this question, you are given the value of Kp and the initial partial pressures of H2 and I2, and you need to determine the equilibrium partial pressure of HI. To solve for the equilibrium partial pressure of HI, you can use the equation Kp = (p(HI))^2 / (p(H2) * p(I2)) and substitute the given values. Solving for p(HI), you can find the equilibrium partial pressure of HI in the sealed reaction vessel at 825 °C.

Answer 2

The pressure of the HI at equilibrium is 0.22 atm

We have that;

Reaction is;

`H_2` +
`I_2` ------> 2HI

I 0.140 0.140 0

C -x -x +2x

E 0.140 - x 0.140 - x 2x

Kp =
`[HI]^2`/[
`H_2` ] [
`I_2` ]

83 =
`(2x)^2`/
`( 0.140 - x )^2`

83 =
`(2x)^2`/
`x^2` - 0.28x + 0.0196

83(
`x^2` - 0.28x + 0.0196) =
`(2x)^2`

`83x^2` - 23.24x + 1.63 =
`4x^2`

`83x^2 - 4x^2 - 23.24x + 1.6 = 0`

x = 0.11( The value can not exceed the initial concentration)

At equilibrium, we have ;

[HI] = 2(0.11)

= 0.22 atm