Respuesta :
Answer : The partial pressure of [tex]H_2[/tex] at equilibrium is, 1.0 × 10⁻⁶
Explanation :
The partial pressure of [tex]HBr[/tex] = [tex]1.0\times 10^{-2}atm[/tex]
The partial pressure of [tex]H_2[/tex] = [tex]2.0\times 10^{-4}atm[/tex]
The partial pressure of [tex]Br_2[/tex] = [tex]2.0\times 10^{-4}atm[/tex]
[tex]K_p=4.2\times 10^{-9}[/tex]
The balanced equilibrium reaction is,
[tex]2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)[/tex]
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 [tex]K_p[/tex] for the reaction will be:
[tex]K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}[/tex]
Now put all the values in this expression, we get :
[tex]4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}[/tex]
[tex]p=-1.99\times 10^{-4}[/tex]
The partial pressure of [tex]H_2[/tex] at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶
Therefore, the partial pressure of [tex]H_2[/tex] at equilibrium is, 1.0 × 10⁻⁶
