Question 22 of 23 Organic Chemistry Roberts & Company Publishers presented by Sa

Question

Question 22 of 23 Organic Chemistry Roberts & Company Publishers presented by Sapling Learming A reactiorn Alaq)+ Blaq)Clag) has a standard free-energy change of 4.81 kJ/mol at 25°C. What are the concentrations of A B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively? Number Number Number How would your answers above change if the reaction had a standard free-energy change of +481 kJimol? O O O There would be less A and B but more C. All concentrations would be higher. All concentrations would be lower. There would be no change to the answers There would be more A and B but less C O OPrevnus @l.nm ilii meiH iuanti @check Answer O NextExit Hint

Explanation / Answer

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)

from dG

dG° = -RT*ln(K)

K = exp(-dG°/(RT))

K = exp(4810/(8.314*298)) = 6.9686

now..

reaction will take place until there is equilibrium achieved.

For this, we use Q, the reaction quotient of products/reactants, it allows us to understand the ratio distribution and the direction/shit of equilibrium

Q is defined as:

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

In this Case, the concentrations are NOT in equilibrium

Therefore:

If Q < Keq; this has much more reactants than products, therefore expect reactants to form more product in order to achieve equilibrium

If Q > Keq; this has much more products than reactants, therefore expect products to form more reactants in order to achieve equilibrium

If Q = Keq; this has the same ratio in equilibrium for reactants and products. Expect no reaction. It is safe to assume this is already in equilibrium.

get Q

Q = [C] /([A][B]) = (0)/(0.3*0.4 = 0

then, this proceeds forward

apply:

initiayl

[A] = 0.3

[B] = 0.4

[C] = 0

in equilbrium

[A] = 0.3 - x

[B] = 0.4 - x

[C] = 0 + x

6.9686 = (x)/((0.4 - x)(0.3 - x))

0.4*0.3 - 0.7x + x^2 = 1/6.9686 x

x^2 + (-0.7-1/6.9686 )x + 0.4*0.3 = 0

x^2 -0.843x +0.12 = 0

x = 0.18137

[A] = 0.3 - 0.18137 = 0.11863

[B] = 0.4 - 0.18137 = 0.21863

[C] = 0 + 0.18137 = 0.18137

Q2.

if dG is positive,

we still have some C formation, since [C] is 0, which is not possible in equilbirium

note that

[C] << [A][B] in this case

choose

there would be more A and B but less C

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