Respuesta :
Answer : The value of equilibrium constant (K) is, 0.11
Explanation :
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.
As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.
The given chemical reaction is:
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
The expression for equilibrium constant is:
[tex]K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]
Now put all the given values in this expression, we get:
[tex]K_c=\frac{(0.47)^2}{(1.5)\times (1.1)^3}[/tex]
[tex]K_c=0.11[/tex]
Thus, the value of equilibrium constant (K) is, 0.11
Answer:
The value of [tex]Kc[/tex] for the reaction [tex]$N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$[/tex] is [tex]0.11[/tex]
Explanation:
Equilibrium concentration is the ratio of products to the density of reactants.
Concentration at equilibrium [tex]\ \ \ N_{2}+3 H_{2} \rightleftharpoons 2 N H_{3}&$\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1.5 \ \ \ 1.1 \ \ \ \ \ \ \ \ 04\end{array}$[/tex]
The given chemical equation is
[tex]$N_{2}(g)+3 H_{2}(g) \rightarrow 2 N H_{3}(g)$[/tex]
The expression of the equilibrium constant, [tex]Kc[/tex] is
[tex]$K_{c}=\frac{\left[N H_{3}\right]^{2}}{\left[N_{2}\right]\left[H_{2}\right]^{3}}$[/tex]
Now substitute all the given values,
we get:
[tex]$K_{c}=\frac{(0.47)^{2}}{(1.5) \times(1.1)^{3}}$[/tex]
[tex]$K_{c}=0.11$[/tex]
Learn more about Equilibrium constant, refer:
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