Q194ptsConsider the chemical reaction 2NHg N2g 3H2g The equ

Q19.(4pts)--Consider the chemical reaction 2NH(g) N2(g) + 3H2(g). The equi libriumis to be established in a 1.0 L container at 1,000 K, where K.-4.0 × 10-2. Initially, 1,220 moles of NHs(g) are present. Estimate the equilibrium concentration of H2(g)

Solution

Initial concentration of NH3=[NH3]o=1220 mol/1.0L=1220mol/L

ICE table

Kc=4.0*10^-2=[N2][H2]^3/[NH3]^2=(x)(3x)^3/(1220-2x)^2

4.0*10^-2=27x^4/(1220-2x)^2

Taking square root of both sides and solving for x,

X=3.0M

[H2]=3*x=9M

	[NH3]	[N2]	[H2]
Initial	1220M	0	0
Change	-2x	+x	+3x
Equilibrium	1220M-2x	x	3x