The equilibrium constant for the reaction N2(g)+3H2(g)2NH3(g) is Kp=4.34×103 at

Question

The equilibrium constant for the reaction N2(g)+3H2(g)2NH3(g) is Kp=4.34×103 at 300 C. Pure NH3 is placed in a 1.00-L flask and allowed to reach equilibrium at this temperature. There are 1.04 g NH3 in the equilibrium mixture.

Part A What is the mass of N2 in the equilibrium mixture? m = g SubmitMy AnswersGive Up

Part B What is the mass of H2 in the equilibrium mixture? m = g SubmitMy AnswersGive Up

Part C What was the initial mass of ammonia placed in the vessel? m = g SubmitMy AnswersGive Up

Part D What is the total pressure in the vessel? Pt = atm

Explanation / Answer

consider the given reaction

N2 + 3H2 ---> 2NH3

Kp = [pNH3]^2 / [pN2] [pH2]^3

now

given

1.04 g of NH3 is present at equilibrium

we know that

moles = mass / molar mass

so

moles of NH3 = 1.04 / 17 = 0.06117647

now

PV = nRT

P x 1 = 0.06117647 x 0.0821 x 573

P = 2.878

so

pNH3 = 2.878

now

initially no N2 and 3H2 are present

so

at equilibrium

pN2 = y

pH2 = 3y

pNH3 = 2.878

so

4.34 x 10-3 = [ 2.878]^2 / [ y] [3y]^3

y = 2.90

so

at equibrium

pN2 = 2.90

pH2 = 3 x 2.90 = 8.7

A)

now

PV = nRT

so

2.9 x 1 = n x 0.0821 x 573

n = 0.061645335

now

mass of N2 = moles x molar mass

so

mass of N2 = 0.061645335 x 28

mass of N2 = 1.726

so

1.726 grams of N2 is present in equilibrium mixture

B)

PV = nRT

8.7 x 1 = n x 0.0821 x 573

n = 0.184936

mass of H2 = 0.184936 x 2

mass of H2 = 0.37 g

so

0.37 g of H2 is present in the equilibrium mixture

C)

now

equilibrium pressure of NH3 = initial pressure - 2y

2.878 = initial pressure - ( 2 x 2.9)

initial pressure of NH3 = 8.678

now

PV = nRT

8.678 x 1 = n x 0.0821 x 573

n = 0.184468

now

mass of NH3 = 0.184468 x 17

mass of NH3 = 3.136

so

3.136 g of NH3 is initially present

D)

now

at equilibtium

total pressure = pN2 + pH2 + pNH3

total pressure = 2.9 + 8.7 + 2.878

total pressure = 14.478

so

total pressure in the vessel is 14.478 atm

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