At 773 K, the equilibrium constant `K_(c)` for the reaction, `N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)" is " 6.02 xx 10^(-2)L^(2) mol^(-2).` Calculate the value of `K_(p)` at the same temperature.

At 773 K, the equilibrium constant `K_(c)` for the reaction, `N_(2) (g

1.	At 773 K, the equilibrium constant `K_(c)` for the reaction, `N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)" is " 6.02 xx 10^(-2)L^(2) mol^(-2).` Calculate the value of `K_(p)` at the same temperature.
Answer» `Delta n_(g) = 2 - 4 = -2, K_p= K_(c) (RT)^(Delta n)` `=6.02 xx 10^(-2) L^(2) mol^(-2) xx (0*0821 L " atm " K^(-1)"mol"^(-1) xx 773 K)^(-2)` ` = 1.5 xx 10^(-5) "atm"^(-2)`

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At 773 K, the equilibrium constant `K_(c)` for the reaction, `N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)" is " 6.02 xx 10^(-2)L^(2) mol^(-2).` Calculate the value of `K_(p)` at the same temperature.

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