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.

Part A

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction

2COF2(g)CO2(g)+CF4(g),    Kc=8.60

If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2remains at equilibrium?

Part B

Consider the reaction

CO(g)+NH3(g)HCONH2(g),    Kc=0.880

If a reaction vessel initially contains only CO and NH3 at concentrations of 1.00 M and 2.00 M, respectively, what will the concentration of HCONH2 be at equilibrium?

Explanation / Answer

A)

consider the given reaction

2COF2 ---> C02 + CF4

initially 2 M of COF2 is present

using ICE table

at equilibrium

[COF2] = 2 - 2x

[C02] = x

[CF4] = x

now

Kc = [C02] [CF4] / [COF2]^2

8.6 = [x] [x] / [2-2x]^2

8.6 = [ x / (2-2x)]^2

2.93258 = x / (2-2x)

2.93258 ( 2-2x) = x

5.86516 - 5.86516x = x

x = 0.85434

so

at equillibirum

[C02] = [CF4] = 0.85434 M

[COF2] = 2 - (2 * 0.85434) = 0.29132 M

B)

CO + NH3 ---> HC0NH2

initially 1 M CO and 2 M NH3 are present

now

using ICE table

at equilibrium

[CO] = 1 - x

[NH3] = 2 - x

[HCONH2] = x

now

Kc = [HCONH2] / [CO] [NH3]

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

0.88 ( x2 - 3x + 2) = x

0.88x2 - 2.64x + 1.76 = x

0.88x2 - 3.64x + 1.76 = 0

x = 0.56

so

at equilibrium

[HCONH2] = x = 0.559 M

[CO] = 1 - 0.559 = 0.441

[NH3] = 2 - 0.559 = 1.441 M

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