for the equilibrium
2HI(g)--- H2(g) + I2 k eq = 8.0
2.0 mol of HI are placed in a 4.0 L container, and the system is allowed to reach equilibrium. Calculate the equilibrium concentration of all three gases.
Inital
2HI 0.5mol/l
H2 =0
I2=0
change
2HI= -2x
H2=+x
I2=+x
equilibrium
2HI = 0.5-2x
H2= x
I2=x
keq= H2 x I2/ HI^2
8.0= (x) (x)/0.5-2x
taking sqr root of both sides
2.828=x/0.5-2x
0.2154=x
so 0.2154 mol/L of H2 and I2 at equilibrium and
2HI = 2x=2(0.2154)= 0.4308
0.5-0.4308=0.069
2HI=0.069 mol/L
If you could look this over to see if I have done this right. I appreciate any help you can give.
Let's go through your calculations to determine if they are correct.
1. You correctly set up the initial concentrations of the gases:
- [HI] = 2.0 mol / 4.0 L = 0.5 mol/L
- [H2] = 0 mol/L
- [I2] = 0 mol/L
2. You correctly set up the change in concentrations for each species at equilibrium:
- Δ[HI] = -2x
- Δ[H2] = +x
- Δ[I2] = +x
3. You correctly set up the equilibrium concentrations for each species:
- [HI]eq = 0.5 - 2x
- [H2]eq = x
- [I2]eq = x
4. You correctly set up the expression for equilibrium constant:
- Keq = [H2]eq * [I2]eq / [HI]eq^2
- Keq = x * x / (0.5 - 2x)^2
5. You correctly substituted the given Keq value and solved for x:
- 8.0 = (x * x) / (0.5 - 2x)^2
- Taking the square root of both sides gives: 2.828 = x / (0.5 - 2x)
6. You correctly solved for x: 0.2154 = x
7. You correctly calculated the equilibrium concentrations:
- [H2]eq = 0.2154 mol/L
- [I2]eq = 0.2154 mol/L
- [HI]eq = 2 * x = 2 * 0.2154 = 0.4308 mol/L
8. However, you made a small mistake in calculating the equilibrium concentration of HI at the end. The correct value is 0.4308 mol/L, not 0.069 mol/L.
So, overall, your calculations are correct, except for the final value of [HI]eq.
Correct equilibrium concentrations:
- [HI]eq = 0.4308 mol/L
- [H2]eq = 0.2154 mol/L
- [I2]eq = 0.2154 mol/L