Respuesta :
According to the balanced equation of this reaction:
N2(g) + 3H2(g) ↔ 2NH3(g)
and when we have Kp = 4.51 x 10^-5 so, in the Kp equation we will substitute by the value of the P for each gas to compare the value with Kp = 4.51x10^-5
a) when we have 98 atm NH3, 45 atm N2, 55 atm H2 by substitution in Kp equation:
Kp= [p(NH3)]^2 / [p(N2)]*[p(H2)]^3 = [98]^2 / [45]*[55]^3
= 1.28x10^-3
So here the value is higher than the value of the given Kp.
so the reaction will go leftwards toward the reactants ( to reduce the value of Kp) to reach the equilibrium.
b) When 57 atm NH3, 143 atm N2, No H2 so like a) by substitution:
Kp = [57]^2 / [143] = 22.7
So the reaction will go leftwards toward the reactants to reduce the value of Kp to reach equilibrium.
c) when 13 atm NH3, 27 atm N2, 82 H2
Kp = [13]^2 / [27]*[82]^3 = 1.135 x 10^-5 So this value is lower than the Kp which is given.
so, the reaction will go towards the right toward the products to increase the value of Kp to reach the equilibrium.
N2(g) + 3H2(g) ↔ 2NH3(g)
and when we have Kp = 4.51 x 10^-5 so, in the Kp equation we will substitute by the value of the P for each gas to compare the value with Kp = 4.51x10^-5
a) when we have 98 atm NH3, 45 atm N2, 55 atm H2 by substitution in Kp equation:
Kp= [p(NH3)]^2 / [p(N2)]*[p(H2)]^3 = [98]^2 / [45]*[55]^3
= 1.28x10^-3
So here the value is higher than the value of the given Kp.
so the reaction will go leftwards toward the reactants ( to reduce the value of Kp) to reach the equilibrium.
b) When 57 atm NH3, 143 atm N2, No H2 so like a) by substitution:
Kp = [57]^2 / [143] = 22.7
So the reaction will go leftwards toward the reactants to reduce the value of Kp to reach equilibrium.
c) when 13 atm NH3, 27 atm N2, 82 H2
Kp = [13]^2 / [27]*[82]^3 = 1.135 x 10^-5 So this value is lower than the Kp which is given.
so, the reaction will go towards the right toward the products to increase the value of Kp to reach the equilibrium.