The equilibrium concentration of A is [tex]\boxed{\frac{1}{3}}[/tex].
The equilibrium concentration of B is [tex]\boxed{\frac{1}{3}}[/tex].
The equilibrium concentration of C is [tex]\boxed{\frac{2}{3}}[/tex].
Further explanation:
Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:
[tex]{\text{A(g)}}+{\text{B(g)}}\rightleftharpoons{\text{C(g)}}+{\text{D(g)}}[/tex]
The equilibrium constant is the constant that relates the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for the general reaction is as follows:
[tex]{\text{K}}=\dfrac{{\left[ {\text{D}}\right]\left[{\text{C}}\right]}}{{\left[{\text{A}} \right]\left[{\text{B}}\right]}}[/tex]
Here, K is the equilibrium constant.
The given reaction is,
[tex]{\text{aA}}\left( g \right)+{\text{bB}}\left( g \right) \rightleftharpoons{\text{cC}}\left( g \right)[/tex]
Here,
A and B are the two reactants.
C is the product formed.
a and b are the stoichiometric coefficients of A and B respectively.
c is the stoichiometric coefficient of C.
The expression of [tex]{{\text{K}}_{\text{c}}}[/tex] for the above reaction is as follows:
[tex]{{\text{K}}_{\text{c}}}=\dfrac{{{{\left[{\text{C}}\right]}^{\text{c}}}}}{{{{\left[{\text{A}} \right]}^{\text{a}}}{{\left[{\text{B}}\right]}^{\text{b}}}}}[/tex] ...... (1)
Here,
[tex]{{\text{K}}_{\text{c}}}[/tex] is the equilibrium constant that is concentration-dependent.
Let the change in concentration at equilibrium is x. Therefore, the concentration of C becomes x at equilibrium. The concentration of A and B become 1-x at equilibrium.
Substitute x for [C] , 1-x for [A] and 0.57-x for [B], 1 for a, 1 for b and 2 for c in equation (1).
[tex]{{\text{K}}_{\text{c}}}=\dfrac{{{{\left[ {\text{x}} \right]}^2}}}{{{{\left[{{\text{1 - x}}} \right]}^{\text{1}}}{{\left[{{\text{1 - x}}}\right]}^{\text{1}}}}}[/tex] ...... (2)
Rearrange equation (2) and substitute 4 for [tex]{{\text{K}}_{\text{c}}}[/tex] to calculate the value of x.
[tex]{{\text{x}}^2}=\dfrac{{{\text{8x}} - 4}}{3}[/tex]
The final quadratic equation is,
[tex]{\text{3}}{{\text{x}}^2}-8{\text{x}}+4=0[/tex]
Solve for x,
[tex]{\text{x}}={\text{2 , }}\dfrac{2}{3}[/tex]
The value of x equal to 2 is not accepted as it would make the equilibrium concentration of A and B negative, which is not possible. So the value of x comes out to be 2/3.
The equilibrium concentration of [C] is equal to 2/3.
The equilibrium concentration of A is calculated as follows:
[tex]\begin{aligned}\left[ {\text{A}}\right]&=1-\frac{2}{3}\\&=\frac{1}{3}\\\end{aligned}[/tex]
The equilibrium concentration of B is calculated as follows:
[tex]\begin{aligned}\left[ {\text{B}}\right]&=1-\frac{2}{3}\\&=\frac{1}{3}\\\end{aligned}[/tex]
So the equilibrium concentrations of A, B and C are 1/3, 1/3 and 2/3 respectively.
Learn more:
1. Calculation of equilibrium constant of pure water at 25°c: https://brainly.com/question/3467841
2. Complete equation for the dissociation of (aq): https://brainly.com/question/5425813
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Equilibrium
Keywords: equilibrium constant, A, B, C, a, b, c, 1, 1, 2, 1/3, 1/3, 2/3, Kc, concentration dependent.
