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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

Respuesta :

superman1987
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.

A) The direction in which the mixture must shift to achieve equilibrium is;

Left Direction

B) The direction in which the mixture must shift to achieve equilibrium is;

Left Direction

C) The direction in which the mixture must shift to achieve equilibrium is;

Right Direction

Chemical Equilibrium Equations

We are given the balanced equation reaction at equilibrium as;

N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)

We are given;

Kp value at equilibrium = 4.51 × 10⁻⁵

A) Formula to find Kp in an equilibrium equation is;

Kp =  [P(Product)]ⁿ/[P(Reactant 1)]ⁿ * [P(Reactant 2)]ⁿ

Where;

n is the coefficient attached to the respective product or reactant

P is the pressure

At 98 atm of NH₃, 45 atm N₂, 55 atm H₂

Thus;

Kp = [P(NH3)]²/ [P(N₂)] × [P(H2)]³

Kp = 98²/(45 × 55³)

Kp = 1.28 × 10⁻³

This calculated Kp value is greater than the given Kp value at equilibrium and thus the mixture is not equilibrium but it will shift to the left direction towards the reactants to achieve equilibrium.

B) At 57 atm NH₃, 143 atm N₂, No H₂

Thus;

Kp = [P(NH₃)]²/ [P(N₂)]

Kp = 57²/143

Kp = 22.7

This calculated Kp value is greater than the given Kp value at equilibrium and thus the mixture is not equilibrium but it will shift to the left direction towards the reactants to achieve equilibrium.

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

Thus;

Kp =  [P(NH₃)]²/ [P(N₂)] × [P(H₂)]³

Kp =  13²/(27 × 82³)  

Kp = 1.14 × 10⁻⁵

This calculated Kp value is less than the given Kp value at equilibrium and thus the mixture is not equilibrium but it will shift to the right direction towards the product to achieve equilibrium.

Read more about Chemical Equilibrium Reactions at; https://brainly.com/question/4289021

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