(60 pts) Haber Process (a) Write the balanced equation for the equilibrium Haber

Question

(60 pts) Haber Process (a) Write the balanced equation for the equilibrium Haber Process that produces ammonia from nitrogen and hydrogen (b) If the Kp for this reaction is 4.52 x 105 at 450°C, indicate for each the following three conditions whether the mixture is at equilibrium, and if it is not at equilibrium, what direction it must shift to achieve equilibrium (if applicable): (G) 105 atm ammonia, 495 atm hydrogen, and 35 atm nitrogen (ii) 35 atm ammonia, 595 atm hydrogen, and 0 atm nitrogen (ii) 26 atm ammonia, 42 atm hydrogen, and 202 atm nitrogern (iv) 0 atm ammonia, 687 atm hydrogen, and 0 atm nitrogen (v) 213 atm ammonia, 45 atm hydrogen, and 76 atm nitrogen

Explanation / Answer

Solution :-

Part A

Balanced equation for the equilibrium process of the Ammonia production is the Habber process is as follows

3H2(g) + N2(g)   <--->   2NH3(g)

Part B

From the given Kp value and the pressure of the each reactant and product of the each trial we can calculate the Qp that is the reaction quotient and compare if the reaction is at the equilibrium state of not and to which direction it will shift to attain the equilibrium

Given Kp = 4.52*10^-5

Qp equation for the reaction is

Qp =[NH3]^2/ [H2]^3[N2]

So lets put the given pressures of the each trial in the Qp equation and calculate the Qp for the each trial

(i) Qp = [105]^2 /[495]^3[35]

          = 2.60*10^-6

Here Qp is smaller than Kp therefore reaction will shift to right side to attain the equilibrium

(ii) Qp = [35]^2 /[595]^3 [0]

Here since the N2 pressure is 0 therefore the reaction is not equilibrium and it will shift to left side (reactant side) to attain the equilibrium

(iii) Qp =[26]^2/[42]^3[202]

            = 4.52*10^-5

Here Qp and Kp are same therefore the reaction is at equilibrium position.

(iv) Qp = [0]^2 /[687]^3[0]

Here since the ammonia (NH3) and N2   pressures are zero therefore it will not shift to any direction.

(v) Qp = [213]^2/[45]^3[76]

            = 6.55*10^-3

Here Qp is greater than Kp therefore reaction will shift to left side to attain the equilibrium.

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