Question

At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures of HBr, H2, and Br2 are 1.0 x 10^-2 atm, 2.0 x 10^-4 atm, and 2.0 x 10^-4 atm, respecivtely, what is the equilbrium partial pressure of H2?

Answer 1

Answer : The partial pressure of
`H_2` at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of
`HBr` =
`1.0* 10^(-2)atm`

The partial pressure of
`H_2` =
`2.0* 10^(-4)atm`

The partial pressure of
`Br_2` =
`2.0* 10^(-4)atm`

`K_p=4.2* 10^(-9)`

The balanced equilibrium reaction is,

`2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)`

Initial pressure 1.0×10⁻² 2.0×10⁻⁴ 2.0×10⁻⁴

At eqm. (1.0×10⁻²-2p) (2.0×10⁻⁴+p) (2.0×10⁻⁴+p)

The expression of equilibrium constant
`K_p` for the reaction will be:

`K_p=((p_(H_2))(p_(Br_2)))/((p_(HBr))^2)`

Now put all the values in this expression, we get :

`4.2* 10^(-9)=((2.0* 10^(-4)+p)(2.0* 10^(-4)+p))/((1.0* 10^(-2)-2p)^2)`

`p=-1.99* 10^(-4)`

The partial pressure of
`H_2` at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of
`H_2` at equilibrium is, 1.0 × 10⁻⁶