For the reaction 2HI(g) <----> H2(g) + I2(g),
Kc = 0.0198 at 721K. What is the HI concentration at 721K if [H2] = 0.120 M and [I2] = 0.0150 M?
I'm getting .303 M but that's not one of the options.
To find the concentration of HI at 721K using the given concentrations of H2 and I2, we can use the equilibrium constant (Kc) expression and apply the concept of stoichiometry.
The balanced chemical equation for the reaction is: 2HI(g) ⇌ H2(g) + I2(g)
The equilibrium constant expression (Kc) for the given reaction is:
Kc = [H2][I2] / [HI]^2
Now, we can rearrange the equation to solve for [HI]:
[HI]^2 = ([H2][I2]) / Kc
Plug in the given values:
[H2] = 0.120 M
[I2] = 0.0150 M
Kc = 0.0198
Substitute the values into the equation:
[HI]^2 = (0.120 M * 0.0150 M) / 0.0198
Calculate:
[HI]^2 = 0.090 M^2 / 0.0198
[HI]^2 ≈ 4.545 M^2
Take the square root of both sides to find [HI]:
[HI] ≈ √(4.545 M^2)
[HI] ≈ 2.134 M
Therefore, the concentration of HI at 721K is approximately 2.134 M.