Question

N₂ + H₂ <=> 2NH₃ Given in part A: Kp = 1.64x10^-4 at 673K and they asked what is the Kc value, I got 5.38x10^-8.

Now they are asking in Part B, qith 105 moles N2 and 318 moles H2 in a 400L gas chamber, how much ammonia would be produced?.... I set up and ICE table with the Es equaling: E N₂= 105-1x, E H₂= 318-3x and E NH₃ = +2x

following I Set up this, Kc=1.64x10^-4(2x)² /(105-1x)(318-3x)³ ....I believe to solve for X or NH₃a quadratic formula is to be used now, how would you solve for this problem/ Set it up?

Answer 1

N2(g) + 3H2(g) <-----> 2NH3(g)

Kp= Kc(RT)^Dn

Dn= moles of gas at products - moles of gas at reagents = 2 - 4 = -2

1.64x10^-4= Kc (0.082L.atm/mol.K x 673K)^-2  ----> Kc= 0.499

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Part B: First we have to transform moles to concentration if we are going to use Kc, or to pressure if we are going to use Kp. I will use Kc, so i will calculate concentrations:

N2(g) + 3H2(g) <-----> 2NH3(g)

105/400L_318/400L_____0

-x______-3x_________2x

0.2625M-x___0.795M-3x______2x

Kc= (2x)²/(0.2625-x)(0.795-3x)³

we can assume that x is a small amount compared with the concentration of N2 and H2 in order to simplify the equation:

Kc=(2x)²/(0.2625)(0.795)³= 0.499 -----> x= 0.181M

[NH3]= 2x= 2(0.181)= 0.363M

they ask how much, but not the unit, i mean, moles or grams? i will calculate both:

mol NH3= 0.363M x 400L= 145 mol ----> mass= 145mol x 17g/mol= 2465g