Answer:
[tex]1.742\times 10^{-5}[/tex] is the value of the equilibrium constant at this temperature.
Explanation:
[tex]2H_2O\rightleftharpoons 2H_2+O_2[/tex]
We are given:
Partial pressure of [tex]H_2O=p^o_{H_2O}=0.0750 atm[/tex]
Partial pressure of [tex]H_2=p^o_{H_2}=0.00700 atm[/tex]
Partial pressure of [tex]O_2=p^o_{O_2}=0.00200 atm[/tex]
The expression of [tex]K_p[/tex] for the given chemical equation is:
[tex]K_p=\frac{p^o_{H_2}^2\times p^o_{O_2}}{p^o_{H_2O}^2}[/tex]
Putting values in above equation, we get:
[tex]K_p=\frac{(0.00700 atm)^2\times 0.00200 atm}{(0.0750 atm)^2}\\\\K_p=1.742\times 10^{-5}[/tex]
[tex]1.742\times 10^{-5}[/tex] is the value of the equilibrium constant at this temperature.
The value of the equilibrium constant at the temperature in which the reaction proceeds is 1.74×10¯⁵
From the question given above, the following data were obtained:
Partial pressure of H₂O (P_H₂O) = 0.0750 atm
Partial pressure of H₂ (P_H₂ ) = 0.00700 atm
Partial pressure of O₂ (P_O₂) = 0.00200 atm
The equilibrium constant for the reaction can be obtained as follow:
2H₂O <==> 2H₂ + O₂
K = (P_H₂)² × (P_O₂) / (P_H₂O)²
K = (0.007)² × (0.002) / (0.075)²
Therefore, the equilibrium constant for the reaction is 1.74×10¯⁵
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