Chemistry, 02.02.2020 15:56 23rwilliamson

Initially, only a and b are present, each at 2.00 m. what is the final concentration of a once equilibrium is reached? kc = 6.5

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Answer from: kandiceboggs1

The final concentration of A after the establishment of equilibrium $\boxed{{\text{0}}{\text{.48309 M}}}$ .

Further Explanation:

Chemical equilibrium is a condition where the rate of forward reaction becomes equal to that of backward reaction. At equilibrium, the formation of product from reactant gets balanced out by the formation of reactants from products so there is no change in concentrations of both reactants and products.

The general equilibrium reaction can be written as follows:

$a{\text{A}} + b{\text{B}} \rightleftharpoons c{\text{C}} + d{\text{D}}$

Here,

A and B are the reactants.

C and D are the products.

a and b are the stoichiometric coefficients of reactants.

c and d are the stoichiometric coefficients of products.

The expression of the equilibrium constant for the general reaction is as follows:

${K_{\text{c}}} = \dfrac{{{{\left[ {\text{C}} \right]}^c}{{\left[ {\text{D}} \right]}^d}}}{{{{\left[ {\text{A}} \right]}^a}{{\left[ {\text{B}} \right]}^b}}}$

Here,

${K_{\text{c}}}$ is the equilibrium constant.

[C] is the concentration of C.

[D] is the concentration of D.

[A] is the concentration of A.

[B] is the concentration of B.

Consider the given reaction,

${\text{A}} + {\text{B}} \rightleftharpoons {\text{AB}}$

The initial concentration of both A and B is 2 M. Let x to be the change in concentration at equilibrium. Therefore, the concentration of AB becomes x at equilibrium. The concentration of both A and B become 2-x at equilibrium.

The expression of ${K_{\text{c}}}$ for the given reaction is as follows:

${K_{\text{c}}} = \dfrac{{\left[ {{\text{AB}}} \right]}}{{\left[ {\text{A}} \right]\left[ {\text{B}} \right]}}$ …… (1)

Substitute x for [AB], 2-x for [A], 2-x for [B] and 6.5 for ${K_{\text{c}}}$ in equation (1).

$6.5 = \dfrac{{\text{x}}}{{\left( {2 - x} \right)\left( {2 - x} \right)}}$ …… (2)

Rearrange equation (2) to form the quadratic equation as follows:

${x^2} - 4.1538x + 4 = 0$

Solve for x,

${\text{x}} = {\text{1}}{\text{.51691 , }}2.63693$

The value of x equal to 2.63693 is rejected as it would make the equilibrium concentration of A and B negative but concentrations cannot be negative. So the value of x is 1.51691.

The concentration of A at equilibrium is calculated as follows:

$\begin{aligned}\left[{\text{A}} \right]&= 2 - 1.51691\\&= 0.48309{\text{ M}} \\\end{aligned}$

Therefore the concentration of A after the establishment of equilibrium is 0.48309 M.

Learn more:

Calculation of equilibrium constant of pure water at : Complete equation for the dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ (aq):

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: chemical equilibrium, reactants, products, concentration, A, B, C, D, a, b, c, d, kc, equilibrium constant, 0.48309 M.

Answer from: Quest

Iam not sure butch i thinnk it is ab

Answer from: Quest

Hey how did you manage to type such a big lab report

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Initially, only a and b are present, each at 2.00 m. what is the final concentration of a once equil...