Answer:
The value of equilibrium constant = 0.005184
Explanation:
According to the given reaction we consider Kp equation and find the value of equilibrium constant using partial pressure of individual gases.
2 H 2 O ( g ) ⇄ 2 H 2 ( g ) + O 2 ( g )
Kp = [tex]\frac{Partial pressure of products}{Partial pressure of reactants}[/tex]
[tex]K_{p} = \frac{P_{H_{2}^{2} }X P{o_{2} } }{P_{H_{2}O^{2} } }[/tex]
Change of gaseous moles(Δn)
= 3 - 2
= 1
⇒ [tex]K_{p} = \frac{(0.006)^{2}X(0.009) }{(0.025)^{2} } \\Kp = 0.005184[/tex]
∴ The value of equilibrium constant = 0.005184
