Answer:
[tex]K_p= K_c\times (RT)^{-3}[/tex]
Explanation:
The relation between Kp and Kc is given below:
[tex]K_p= K_c\times (RT)^{\Delta n}[/tex]
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant , R = 0.082057 L atm.mol⁻¹K⁻¹
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
[tex]2H_2_{(g)}+O_2_{(g)}\rightleftharpoons2H_2O_{(l)} [/tex]
Δn = (0)-(2+1) = -3
Thus, Kp is:
[tex]K_p= K_c\times (RT)^{-3}[/tex]