Could I get help solving this problem? What I have so far is that it\'s an equil
ID: 929600 • Letter: C
Question
Could I get help solving this problem? What I have so far is that it's an equilibrium problem so I should set up an ICE table. I can use the 1.05g NH3 at equilibrium to calculate the moles of NH3 at equilibrium (using molar mass). Is that right? I just don't get how i will get th einitial concentration of NH3 and the next steps.
The equilibrium constant for the reaction:
N2 (g) + 3 H2 (g) 2 NH3 (g) is 4.34 x 10-3 at 300oC.
Pure NH3 is placed in a 1.00 L flask and allowed to reach equilibrium at this temperature.
There are 1.05 g NH3in the equilibrium mixture.
What are the masses of N2 and H2 in the equilibrium mixture?
What was the initial mass of ammonia placed in the vessel?
What is the total pressure in the vessel?
Explanation / Answer
at equilibrium NH3 moles = a-X where a = initial moles = mass of Nh3 / molar massof NH3
= 1.05/17.031 = 0.0616523
N2 moles = X , H2 moles = 3X
K = [Nh3]^2 /[N2][H2]^3
0.00434 = ( 0.0616523) /( X) ( 3X)^3
27X^4 = ( 0.0616523/0.00434)
X = 0.851675
now N2 moles = 0.851675 , N2 mass = moles x molar mass of N2 = 0.851675 x 28 = 23.85 g
H2 moles = 3X = 3( 0.851675) = 2.555 , H2 mass = 2.555 x 2 = 5.1 g
Nh3 equilibrium moles = a-X = a-0.851675 = 0.0616523
a = initia NH3 moles = 0.9133
NH3 mas s= 0.9133 x 17.031 = 15.555 g
Now total gas moles at equilibrium = 0.0616523+0.851675 + 2.555 = 3.4683
we use PV = nRT eqution toget total P , T = 300C = 300+273 = 573 K
P x 1 = 3.4683 x0.08206 x 573
P = 163 atm