nicolepavlovsky
contestada

A student ran the following reaction in the laboratory at 636 K:

2HI(g) -> H2(g) + I2(g)

When she introduced 0.316 moles of HI(g) into a 1.00 liter container, she found the equilibrium concentration of I2(g) to be 3.12×10-2 M.

Calculate the equilibrium constant, Kc, she obtained for this reaction.

Kc =

Respuesta :

Edufirst

Answer:

      [tex]\large\boxed{\large\boxed{K_c=1.51\times 10^{-2}}}[/tex]

Explanation:

1. Calculate the iniital concentration of HI(g) introduced

The initial concentration of HI(g), [HI(g)], is the number of moles per liter:

  • [HI(g)] = 0.316mol/1liter = 0.316M

2. Build the ICE (initial, change, equilibrium) table

ICE table:

          2HI(g)          →        H₂(g)      +       I₂(g)

I          0.316                         0                   0

C         - 2x                         +x                   +x      

E         0.316 - 2x                 x                    x       (x =  3.12×10⁻²M)

3. Find the equilibrium constant, Kc

           [tex]K_c=\dfrac{x\cdot x}{(0.316-2x)^2}\\\\\\K_c=\dfrac{(3.12\times 10^{-2}M)^2}{\big(0.316M-2(3.12\times 10^{-2}M)\big)^2}\\\\\\K_c=1.51\times 10^{-2}[/tex]

The equilibrium constant for the reaction will be: [tex]K_c=1.51*10^{-2}[/tex]

Calculation for initial concentration of HI(g) produced:

The initial concentration of HI(g), [HI(g)], is the number of moles per liter:

[HI(g)] = 0.316mol/1liter = 0.316M

Chemical reaction:

         2HI(g)          →        H₂(g)      +       I₂(g)

I          0.316                         0                   0

C         - 2x                         +x                   +x      

E         0.316 - 2x                 x                    x       (x =  3.12×10⁻²M)

Calculation for equilibrium constant:

[tex]K_c=\frac{x.x}{(0.316-2x)^2} \\\\K_c=\frac{(3.2*10^ {-2})^2}{(0.316-2*3.12*10^{-2})^2}\\\\K_c=1.51*10^{-2}[/tex]

Find more information about Equilibrium constant here:

brainly.com/question/12858312          

Otras preguntas