The concentrations of reactants and products for a chemical reaction can be calc

Question

The concentrations of reactants and products for a chemical reaction can be calculated if the equilibrium constant for the reaction and the starting concentrations of reactants and/or products are known. Carbonyl fluoride, COF_2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF-i via the reaction 2COF_2 (g) CO_2 (g) + CF_4 (g), Kc = 7.80 If only COF_2 is present initially at a concentration of 2.00 M, what concentration of COF_2 remains at equilibrium? Express your answer with the appropriate units.

Explanation / Answer

initially

[COF2] = 2

[CO2] = 0

[CF4] = 0

in equilibrium

[COF2] = 2 -2x

[CO2] = 0+x

[CF4] = 0+x

then

K = [CO2][CF4]/([COF2]^2)

K = 7.8

7.8 = (x)(x)/(2-2x)^2

solve for x

sqrt(7.8) = x/(2-2x)

2.79*(2-2x)= x

2-2x = 0.358x

(2+0.358)x = 2

x = 2/((2+0.358) = 0.84817

then

[COF2] = 2 -2x = 2-2*0.84817= 0.30366

[CO2] = 0+x = 0.84817

[CF4] = 0+x = 0.84817

JUST TO MAKE SURE

test in the Kc expression

it must satisfy

Kc = 7.80

then

K = [CO2][CF4]/([COF2]^2)

K = (0.84817)(0.84817)/(0.30366^2) = 7.8017

it is pretty accurate

2)

[CO] = 1

[NH3] = 2

[urea] = 0

in equilbirium

[CO] = 1-x

[NH3] = 2-x

[urea] = 0+x

then

K = [urea]/[CO][NH3]

0.88 = (x)/(1-x)(2-x)

(2-3x+x^2) = 1/0.88 x

x^2 - 4.13x + 2 = 0

x = 0.56

[urea] = 0+x = 0.56 M

TEST NOW FOR Kc

Kc = 0.88

then

K = [urea]/[CO][NH3]

K = (0.56 )/(1-0.56 )/(2-0.56 ) = 0.883838

PRETTY ACCURATE

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