472℃ 7.38 atm H2, 2.46 atm N2, 0.166 atm NH3

472℃ 7.38 atm H2, 2.46 atm N2, 0.166 atm NH3

472℃ 7.38 bar H2, 2.46 bar N2, 0.166 bar NH3

After a mixture of hydrogen and nitrogen gases in a reaction vessel

is allowed to attain equilibrium at 472℃,

it is found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3.

From these data,

calculate the equilibrium constant, Kp, for the reaction

N2(g) + 3H2(g) → 2NH3(g).

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N2(g) + 3H2(g) → 2NH3(g)

Kp = (P_NH3)^2 / [(P_N2) (P_H2)^3]

( 참고 https://ywpop.tistory.com/7136 )

= (0.166)^2 / [(2.46) (7.38)^3]

= 2.7868×10^(-5)

답: 2.79×10^(-5)

[키워드] N2 + 3H2 → 2NH3 Kp 기준