2HI(g)--H2(g) + I2(g) k eq= 8.0
2.0 mol of HI are placed in a 4.0L container, and the system is allowed to reach equilibrium. Calculate the equilibrium concentration of all three gases
Initial
2HI=0.5 mol/L
H2= 0
I2=0
change
2HI= -2x
H2= x
I2=x
equilibrium
2HI= 0.5-2x
H2= x
I2=X
k eq= product/reactants
8.0= (x)(x)/ 0.5-2x
2.8=x/0.5-2x
x=0.21
so 0.21 mol/L of H2 and I2
HI= 0.42 0.5-2(0.21) = 0.08mol/L
tough stuff...
The denominator, you need a square
(.5-2x)^2
other wise, right process.
To calculate the equilibrium concentration of all three gases in the given reaction, we need to use the equilibrium constant (K) and the stoichiometry of the reaction.
The given reaction is: 2HI(g) ⇌ H2(g) + I2(g)
According to the stoichiometry, the molar ratio of HI to H2 and I2 is 2:1:1. Therefore, if x moles/L of H2 and I2 are formed at equilibrium, then 2x moles/L of HI would have reacted.
At equilibrium, the initial concentration of HI is given as 0.5 mol/L. Since 2x moles/L of HI reacts, the equilibrium concentration of HI can be calculated as follows:
Equilibrium concentration of HI = Initial concentration of HI - 2x
= 0.5 - 2(0.21)
= 0.5 - 0.42
= 0.08 mol/L
The equilibrium concentrations of H2 and I2 can be directly calculated as x, which is 0.21 mol/L in this case.
Therefore, the equilibrium concentrations of all three gases are:
HI = 0.08 mol/L
H2 = 0.21 mol/L
I2 = 0.21 mol/L