(4) Equilibrium Concentrations -- A + B = 2C At a particular temperature, K = 1.
ID: 593931 • Letter: #
Question
(4) Equilibrium Concentrations -- A + B = 2C
At a particular temperature, K = 1.00×102 for the reaction:
In an experiment, at this temperature, 1.80 mol of H2 and 1.80 mol of F2 are introduced into a 1.50-L flask and allowed to react. At equilibrium, all species remain in the gas phase.
a) What is the equilibrium concentration (in mol/L) of H2?
b) What is the equilibrium concentration (in mol/L) of HF?
c) To the mixture above, an additional 5.04×10-1 mol of H2 is added.
What is the new equilbrium concentration (in mol/L) of HF?
Explanation / Answer
Concentration (M)= moles/ Volume in L
Preparing the ICE Table ( x is the change in concentration of H2 ( or F2 to reach equilibrium)
H2 (M) F2 (M) HF(M)
Initial 1.8/1.5= 1.2 1.8/1.5= 1.2 0
Change -x -x 2x
Equilibrium 1.2-x 1.2-x 2x
Equilibrium constant (K) for the reaction H2(g)+F2(g)<----> 2HF(g)
K= [HF]2/ [H2][F2]= 100
4x2/(1.2-x)2= 100
Taking square root 2x/(1.2-x)= 10, 2x= 10*(1.2-x)= 2x= 12-10x
12x=12, x=1 at Equilibrium [H2]= [F2]=1.2-1=0.2M and [HF]= 2M
When 0.504 moles of H2 is added, new concentrations are
[H2]=0.2+0.504/1.5=0.536, [F2]=0.2M and [HF]=2M
Reaction quotient = [HF]2/ [H2][F2]= 2*2/(0.536*0.2)=37.31<K
So the reaction shift toward the product side
At the new equilibrium, [H2]=0.536-x, and [F2]=0.2-x and [HF]= 2+2x
Where x= change in concentration to reach equilibrium
(2+2x)2/ (0.536-x)*(0.2-x)= 100
When solved using excel,x=0.0924
Hence at new equilibrium, [H2]=0.536-0.0924=0.4436M, [F2]=0.2-0.0924=0.1076M and [HF]= 2.1848M
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