Question

In: Chemistry

An equilibrium mixture of H2, I2, and HI at 458 ∘C contains 0.112 molH2, 0.112 molI2,...

An equilibrium mixture of H2, I2, and HI at 458 ∘C contains 0.112 molH2, 0.112 molI2, and 0.775 molHI in a 5.00-L vessel.

1)What are the equilibrium partial pressure of

HI when equilibrium is reestablished following the addition of 0.200 mol of HI?

Express your answer to four significant figures and include the appropriate units

2)What are the equilibrium partial pressure of

I2 when equilibrium is reestablished following the addition of 0.200 mol of HI?

Express your answer to three significant figures and include the appropriate units.

3) What are the equilibrium partial pressure of

H2 when equilibrium is reestablished following the addition of 0.200 mol of HI?

Express your answer to three significant figures and include the appropriate units.

Solutions

Expert Solution

    H2(g) + I2(g) <---> 2HI(g)

at equilibrium,

concentration of H2 = 0.112/5 = 0.0224 M

concentration of I2 = 0.112/5 = 0.0224 M

concentration of HI = 0.775/5 = 0.155 M

Kc = 0.155^2/(0.0224*0.0224) = 47.88

concentration of HI added = 0.2/5 = 0.04 M

total concentration of HI = 0.155+0.04 = 0.195 M

47.88 = (0.195-2x)/(0.0224+x)^2

x = 0.0305

1)concentration of HI after reestablishing equilibrium

           = 0.195-(2*0.0305)

   = 0.134 M

partial pressure of HI = nRT/V

                        = 0.134*0.0821*(458+273.15)

                        = 8.044 atm


2) concentration of I2 after reestablishing equilibrium

           = 0.0224+0.0305

   = 0.0529 M

partial pressure of I2 = nRT/V

                        = 0.0529*0.0821*(458+273.15)

                        = 3.17 atm

3) concentration of H2 after reestablishing equilibrium

           = 0.0224+0.0305

   = 0.0529 M

partial pressure of I2 = nRT/V

                        = 0.0529*0.0821*(458+273.15)

                        = 3.17 atm


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