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 reaction2COF2(g)⇌CO2(g)+CF4(g), Kc=5.70If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?

[COF₂] = 0.346M

Explanation:

For the reaction:

2COF₂(g) ⇌ CO₂(g) + CF₄(g)

Kc = 5.70 is defined as:

Kc = [CO₂] [CF₄] / [COF₂]² = 5.70 (1)

Equilibrium concentrations of each compound after addition of 2.00M COF₂ will be:

[COF₂] : 2.00M - 2x

[CO₂] : x

[CF₄] : x

Replacing in (1):

5.70 =  [X] [X] / [2-2x]²

22.8 - 45.6x + 22.8x² = x²

0 = -21.8x² + 45.6x - 22.8

Solving for x:

X = 1.265 -False answer, will produce negative concentrations-

X = 0.827.

Replaing, molar concentration of COF₂ is:

[COF₂] : 2.00M - 2×0.827 = 0.346M

I hope it helps!

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

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