Respuesta :
This is an incomplete question, here is a complete question.
Kp = 0.0198 at 721 K for the reaction:
[tex]2HI(g)\rightleftharpoons H_2(g)+I_2(g)[/tex]
In a particular experiment, the partial pressures of H₂ and I₂ at equilibrium are 0.710 and 0.888 atm, respectively. The partial pressure of HI is __________ atm.
Answer : The partial pressure of HI is, 5.64 atm
Explanation :
For the given chemical reaction:
[tex]2HI(g)\rightleftharpoons H_2(g)+I_2(g)[/tex]
The expression of [tex]K_p[/tex] for above reaction follows:
[tex]K_p=\frac{P_{H_2}\times P_{I_2}}{(P_{HI})^2}[/tex]
We are given:
[tex]P_{H_2}=0.710atm[/tex]
[tex]P_{H_2}=0.888atm[/tex]
[tex]k_p=0.0198[/tex]
Now put all the given values in above equation, we get:
[tex]0.0198=\frac{0.710\times 0.888}{(P_{HI})^2}[/tex]
[tex]P_{HI}=5.64atm[/tex]
Thus, the partial pressure of HI is, 5.64 atm
