Answer:
[I₂] ; [H₂] = 0.067 mol/L
[HI] = 0.765 mol/L
Explanation:
This is the reaction:
I₂(g) + H₂(g) ⇄ 2HI (g)
Initially 1mol 1mol 2.5mol
I have the initially amount of each gas at first. Some amount (x), has reacted during the reaction.
I₂(g) + H₂(g) ⇄ 2HI (g)
React x x 2x
In the equilibrium I have to subtract, what I had initially and the amount that has reacted, and then, as I had 2.5 mol of HI at the begining, I have to sum, the amount of the reaction. As I have to find out the concentrations, I have to /5L, which is the volume of the vessel.
Eq. (1-x)/5 (1-x)/5 (2.5+2x)/5
Let's make the expression for Kc
Kc = [HI]² / ( [I₂] . [H₂])
129 = ( (2.5+2x) /5 )² / ( (1-x) /5) . ( (1-x) /5) )
129 = ( (2.5+2x) /5 )² / ( (1-x) /5)²
129 = (2.5+2x)² / 25 / (1-x)² / 25
129 = (2.5 + 2x)² / (1-x)²
129 = 2.5² + 2 . 2.5 .2x + 4x² / 1 - 2x + x²
129 (1 - 2x + x²) = 2.5² + 2 . 2.5 .2x + 4x²
129 - 258x + 129x² = 6.25 + 10x + 4x²
129 - 6.25 -258x -10x + 129x²-4x² = 0
122.75 - 268x + 125x² = 0 (a quadratic function)
a = 125
b = -268
c = 122.75
(-b +- (√b² - 4ac)) / 2a
x₁ = 1.48
x₂ = 0.66
We take the x₂ value, cause the x₁, will get a negative concentration and that is impossible.
[I₂] = ( 1-0.66 ) /5 = 0.067
[H₂] = ( 1-0.66 ) /5 = 0.067
[HI] = (2.5+ 2. 0.66) /5 = 0.765
The equilibrium concentration of H2, I2 and HI are 0.01 M, 0.31 M, 0.19 M.
Molarity of H2 = 1.00mol/5 L = 0.2 M
Molarity of I2 = 2.50mol /5L = 0.5 M
We can set up the ICE table as follows;
H2 + I2 ⇄ 2HI
I 0.2 0.5 0
C -x -x x
E 0.2 - x 0.5 -x x
Kc = [HI]^2/[H2] [I2]
129 = x^2/(0.2 - x) (0.5 -x)
129 = x^2/0.1 - 0.2x - 0.5x + x^2
129(0.1 - 0.2x - 0.5x + x^2) = x^2
12.9 -90.3x + 129x^2 = x^2
x^2 - 129x^2 +90.3x - 12.9 = 0
-128x^2 +90.3x - 12.9 = 0
x = 0.19 M
At equilibrium;
H2 = 0.2 - 0.19 = 0.01 M
I2 = 0.5 - 0.19 = 0.31 M
HI = 0.19 M
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