Answer:
6.75 × 10⁻⁸is the value of the equilibrium constant at this temperature.
Explanation:
2H₂O(g) ⇄ 2H₂(g) + O₂(g)
Partial pressure of H₂O = 0.0500 atm
Partial pressure of H₂ = 0.00150 atm
Partial pressure of O₂ = 0.00150 atm
The expression of Kp for the given chemical equation is:
[tex]K_p = \frac{[H_2]^2[O_2]}{H_2O}[/tex]
[tex]= \frac{(0.00150^2)(0.00150)}{(0.0500)} \\= 6.75 * 10^-^8[/tex]
6.75 × 10⁻⁸is the value of the equilibrium constant at this temperature
