Given data: Kc = 54.3 and the initial concentrations of H2 and I2 are both 0.11 M. H2(g)+I2(g)↽−−⇀2HI(g), Determine the initial and equilibrium concentration of HI.


Find: Determine the initial and equilibrium concentration of HI.

Explanation:

To solve this problem, we can use the equilibrium expression and the given equilibrium constant (Kc) to determine the initial and equilibrium concentrations of HI.

 

Step 2 of 3

The equilibrium expression for the reaction is:

Kc=[HI]2[H2]×[I2]

 

substitute these values into the equilibrium expression to find the initial concentration of HI ([HI]i).

54.3=([HI]i)20.11×0.11

 

Rearranging the equation, we can solve for [HI]i:

[HI]i2=54.3(0.11×0.11)[HI]i2=0.6693[HI]i=√0.6693[HI]i=0.819M

 

Explanation:

Therefore, the initial concentration of HI is 0.819 M.

 

Step 3 of 3

Given that the equilibrium concentrations of H2 and I2 are both 0.042 M,

substitute these values into the equilibrium expression to find the equilibrium concentration of HI ([HI]e).

54.3=[HI]e20.042×0.042

 

Rearranging the equation, we can solve for [HI]e:

[HI]e2=54.3(0.042×0.042)[HI]e2=0.0967[HI]e=√0.0967[HI]e=0.311M

 

Explanation:

Therefore, the equilibrium concentration of HI is 0.311 M.

 

Final solution

The initial and equilibrium concentration of HI is 0.819M and 0.311M.

 

 

 

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