Answer:
[HI] = 0.704mol
[H2] = 0.048mol
[I2] = 0.148mol
Explanation:
Based on the equilibrium:
H2(g)+I2(g) --> 2HI(g)
The equilibrium constant, K, is defined as:
K = 70 = [HI]² / [H2] [I2]
Where [] could be taken as the moles in equilibrium of each reactant
To know the direction of the equilibrium we need to find Q with the initial moles of each species:
Q = [0.200mol]² / [0.300mol] [0.400mol]
Q = 0.333
As Q < K, the reaction will shift to the right producing more HI. The equilibrium moles are:
[HI] = 0.200mol + 2X
[H2] = 0.300mol - X
[I2] = 0.400mol - X
Replacing in K:
70 = [0.200 + 2X ]² / [0.300 - X] [0.400 - X]
70 = 0.04 + 0.8 X + 4 X² / 0.12 - 0.7 X + X²
8.4 - 49 X + 70 X² = 0.04 + 0.8 X + 4 X²
8.36 - 49.8X + 66X² = 0
Solving for X:
X = 0.252 moles. Right solution
X = 0.502 moles. False solution. Produce negative moles.
Replacing:
[HI] = 0.200mol + 2*0.252 mol
[H2] = 0.300mol - 0.252 mol
[I2] = 0.400mol - 0.252 mol
[HI] = 0.704mol
[H2] = 0.048mol
[I2] = 0.148mol
