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
To calculate the equilibrium concentrations of the three gases (HI, H2, and I2) in the given reaction, you can use the information provided and the equilibrium constant (Keq).
First, let's determine the initial concentrations of the gases. It is given that 2.0 mol of HI is placed in a 4.0L container, so the initial concentration of HI is 2.0 mol / 4.0 L = 0.5 mol/L. The initial concentrations of H2 and I2 are both 0 mol/L.
Next, let's determine the change in concentrations of the gases at equilibrium. According to the balanced equation, for every 2 moles of HI that react, you will form 1 mole each of H2 and I2. Therefore, the change in concentration for HI will be -2x, and the change in concentration for both H2 and I2 will be x.
Now, let's express the concentrations of the gases at equilibrium in terms of the change in concentrations and the initial concentrations. The equilibrium concentration of HI will be the difference between the initial concentration and the change in concentration, which is 0.5 mol/L - 2x. The equilibrium concentration of H2 and I2 will be x.
We can now use the equilibrium constant (Keq) expression to relate the concentrations of the gases. The expression for this reaction is: Keq = [H2][I2] / [HI]^2. Substituting the given value of Keq = 8.0 and the equilibrium concentrations we determined, we have:
8.0 = (x)(x) / (0.5 - 2x)^2
You can now solve this equation for x using algebraic manipulations. Once you determine the value of x, you can substitute it back into the equilibrium concentration expressions to obtain the equilibrium concentrations:
[H2] = x
[I2] = x
[HI] = 0.5 - 2x
In this case, x is found to be 0.21 mol/L. Therefore, the equilibrium concentrations are:
[H2] = 0.21 mol/L
[I2] = 0.21 mol/L
[HI] = 0.5 - 2(0.21) = 0.08 mol/L
So, the equilibrium concentrations for the three gases are approximately:
[H2] = 0.21 mol/L
[I2] = 0.21 mol/L
[HI] = 0.08 mol/L