Respuesta :
The equilibrium concentration of NO is [tex]\boxed{{\text{0}}{\text{.0018 M}}}[/tex].
Further Explanation:
Chemical equilibrium is a stage where the rate at which forward reaction proceeds becomes equal to the rate at which the backward reaction occurs. 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 expression for a general equilibrium is,
[tex]a{\text{A}} + b{\text{B}} \rightleftharpoons c{\text{C}} + d{\text{D}}[/tex]
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 for the equilibrium constant for the general reaction is as follows:
[tex]{K_{\text{c}}}=\dfrac{{{{\left[ {\text{C}} \right]}^c}{{\left[ {\text{D}} \right]}^d}}}{{{{\left[ {\text{A}} \right]}^a}{{\left[ {\text{B}} \right]}^b}}}[/tex]
Here,
[tex]{K_{\text{c}}}[/tex] 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.
The given reaction occurs as follows:
[tex]2{\text{NO}}\left( g \right) \rightleftharpoons {{\text{N}}_2}\left( g \right) + {{\text{O}}_{\text{2}}}\left( g \right)[/tex]
The initial concentration of NO is 0.175 M. Let x to be the change in concentration at equilibrium. Therefore, the concentration of NO becomes 0.175-2x at equilibrium. The concentration of both [tex]{{\text{N}}_{\text{2}}}[/tex] and [tex]{{\text{O}}_{\text{2}}}[/tex] become x at equilibrium.
The expression of [tex]{K_{\text{c}}}[/tex] for the above reaction is as follows:
[tex]{K_{\text{c}}}=\dfrac{{\left[ {{{\text{N}}_2}} \right]\left[{{{\text{O}}_2}} \right]}}{{{{\left[ {{\text{NO}}} \right]}^2}}}[/tex] …… (1)
Substitute x for [tex]{{\text{N}}_{\text{2}}}[/tex], x for [tex]{{\text{O}}_{\text{2}}}[/tex], 0.175-2x for [NO] and [tex]2.4 \times {10^3}[/tex] for [tex]{K_{\text{c}}}[/tex] in equation (1).
[tex]2.4 \times {10^3} = \dfrac{{\left( {\text{x}}\right)\left({\text{x}} \right)}}{{{{\left( {0.175 - 2x}\right)}^2}}}[/tex] …… (2)
Solve for x,
[tex]{\text{x}} = 0.0866[/tex]
Or,
[tex]{\text{x}} = 0.0884[/tex]
The value of [tex]{\text{x}} = 0.0884[/tex] is rejected as it makes concentration of NO negative that is not possible. So the value of x is 0.0866.
The concentration of NO at equilibrium is calculated as follows:
[tex]\begin{aligned}\left[ {{\text{NO}}} \right]&= 0.175 - 2\left( {0.0866} \right)\\&= {\text{ 0}}{\text{.0018 M}}\\\end{aligned}[/tex]
Learn more:
- Calculate equilibrium constant for ammonia synthesis: https://brainly.com/question/8983893
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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, 0.0018 M, NO, N2, O2.
