ind the approximate concentrations of each species at equilibrium in the system.

Question

ind the approximate concentrations of each species at equilibrium in the system. Kc = 0.090 0.55 M 0.23 M 0.055 M alculate the equilibrium constant for the reaction shown with equilibrium pressures hown? (Note: you must write out the equilibrium expression first). N2(g) 021g) 2NO(g) + E 0.7609 atm 0.1609 atm 0.07834 atm Convert the Ke 0.0059 value into the Kp value using the following relationship and the eaction below at a temperature of 298 Kelvin: Kp = Kc (RT) an where R = 0.082058 L'atm / mol-K N204 (g) 2 NO2 (g) Kc 0.0059

Explanation / Answer

Q1

initially

H2O = 0.55

Cl2O = 0.23

HOCl = 0.055

the change ( extent of reaciton)

H2O = -x

Cl2O = -x

HOCl = +2x

the equilbirium

H2O = 0.55 - x  

Cl2O = 0.23 - x

HOCl = 0.055 +2x

substitute in K

K = [HOCl]^2 /([H2O][Cl2O])

0.09 = (0.055+2x)^2 / ((0.55 - x)(0.23 - x))

0.09 (0.55*0.23 - (0.23+0.55)x + x^2) = 0.055^2 + 2*0.055x + 4x^2

0.011385 -0.0702x + 0.09x^2 = 0.003025 + 0.11x + 4x^2

(4-0.09)*x^2 + (0.11 + 0.0702)x + (0.003025 -0.011385 ) =0

x = 0.0381

H2O = 0.55 - 0.0381 = 0.5119

Cl2O = 0.23 - 0.0381 = 0.1919

HOCl = 0.055 +2*0.0381 = 0.1312

Q2.

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Kp = PNO^2 / (PN2)(PO2)

kp = (0.07834^2) / (0.7609*0.1609)

Kp = 0.0501

Q3.

substitute data

Kp = Kc*(RT)^dn

Kp = (0.0059)(0.082*298)^(2-1)

Kp =0.1441724

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