As shown in table 15.2, kp for the equilibrium n21g2 + 3 h21g2 δ 2 nh31g2 is 4.51 * 10-5 at 450 °c. for each of the mixtures listed here, indicate whether the mixture is at equilibrium at 450 °c. if it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) 98 atm nh3, 45 atm n2, 55 atm h2 (b) 57 atm nh3, 143 atm n2, no h2 (c) 13 atm nh3, 27 atm n2, 82 atm h2

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Taniya

Chemistry

5 October, 12:05

As shown in table 15.2, kp for the equilibrium n21g2 + 3 h21g2 δ 2 nh31g2 is 4.51 * 10-5 at 450 °c. for each of the mixtures listed here, indicate whether the mixture is at equilibrium at 450 °c. if it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) 98 atm nh3, 45 atm n2, 55 atm h2 (b) 57 atm nh3, 143 atm n2, no h2 (c) 13 atm nh3, 27 atm n2, 82 atm h2

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Gates · Answer 1

Our reaction balanced equation at equilibrium N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)

and we have the Kp value at equilibrium = 4.51 X 10^-5

A) 98 atm NH3, 45 atm N2, 55 atm H2

when Kp = [P (NH3) ]^2 / [P (N2) ] * [P (H2) ]^3

= 98^2 / (45 * 55^3) = 1.28 x 10^-3

by comparing the Kp by the Kp at equilibrium (the given value) So,

Kp > Kp equ So the mixture is not equilibrium,

it will shift leftward (to decrease its value) towards the reactants to achieve equilibrium.

B) 57 atm NH3, 143 atm N2, no H2

∴ Kp = [P (NH3) ]^2 / [P (N2) ]

= 57^2 / 143 = 22.7

∴Kp> Kp equ (the given value)

∴it will shift leftward (to decrease its value) towards reactants to achieve equilibrium.

c) 13 atm NH3, 27 atm N2, 82 atm H2

∴Kp = [P (NH3) ]^2 / [P (N2) ] * [P (H2) ]^3

= 13^2 / (27 * 82^3) = 1.14 X 10^-5

∴ Kp< Kp equ (the given value)

∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.