
Equilibrium constant for a reaction ${K_C}$ is equal to $\dfrac{{{{\left[ {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right]}^4}{{\left[ {{{\rm{O}}_{\rm{2}}}} \right]}^5}}}{{{{\left[ {{\rm{NO}}} \right]}^4}{{\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]}^6}}}$
Answer
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Hint: We know that equilibrium represents a state in a process where there are no changes in the observable properties such as temperature, colour, pressure, temperature etc. In case of physical process the equilibrium is termed as physical equilibrium and in case of chemical reactions the equilibrium is termed as chemical equilibrium.
Complete step by step solution:
Let’s understand the law of chemical equilibrium first. The law states that for a reversible reaction in a state of equilibrium, the ratio of molar concentration of the products to that of molar concentration of reactants; each term raised to a power equal to its stoichiometric coefficient in the balanced chemical reaction is a constant termed as equilibrium constant if temperature is constant.
Consider a reaction,
A+B\rightleftharpoons C+D
The equilibrium constant $\left( {{K_C}} \right)$ for the reaction is,
${K_C} = \dfrac{{\left[ {{\rm{Product}}} \right]}}{{\left[ {{\rm{Reactant}}} \right]}}$
${K_C} = \dfrac{{\left[ {\rm{C}} \right]\left[ {\rm{D}} \right]}}{{\left[ {\rm{A}} \right]\left[ {\rm{B}} \right]}}$
Now, come to the question. Here, the given expression of equilibrium constant is $\dfrac{{{{\left[ {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right]}^4}{{\left[ {{{\rm{O}}_{\rm{2}}}} \right]}^5}}}{{{{\left[ {{\rm{NO}}} \right]}^4}{{\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]}^6}}}$.
We know that the numerator in the equilibrium constant expression represents concentration of products and the denominator represents concentration of the reaction. So, the reactants are NO and ${{\rm{H}}_{\rm{2}}}{\rm{O}}$ and products are \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\] and ${{\rm{O}}_{\rm{2}}}$. The powers in the molar concentration of reactants or products indicate the stoichiometric coefficient in the balanced chemical equation. So, the balanced chemical equation is,
$4NO+6H_{2}O\rightleftharpoons 4NH_{3}+5O_{2}$
Note: If the chemical reaction is in gaseous phase, that is, reactants and products are all gases, then we can also consider the partial pressure of the species involved instead of their molar concentration. In such conditions, the equilibrium constant is denoted by ${K_p}$
.
For the reaction,
$aA+bB\rightleftharpoons cC+dD$
${K_p} = \dfrac{{p{C^c}.p{D^d}}}{{p{A^a}.p{B^b}}}$
Where, $pC$, $pD$, $pA$ and $pB$ are partial pressures of C,D, A and B respectively.
Complete step by step solution:
Let’s understand the law of chemical equilibrium first. The law states that for a reversible reaction in a state of equilibrium, the ratio of molar concentration of the products to that of molar concentration of reactants; each term raised to a power equal to its stoichiometric coefficient in the balanced chemical reaction is a constant termed as equilibrium constant if temperature is constant.
Consider a reaction,
A+B\rightleftharpoons C+D
The equilibrium constant $\left( {{K_C}} \right)$ for the reaction is,
${K_C} = \dfrac{{\left[ {{\rm{Product}}} \right]}}{{\left[ {{\rm{Reactant}}} \right]}}$
${K_C} = \dfrac{{\left[ {\rm{C}} \right]\left[ {\rm{D}} \right]}}{{\left[ {\rm{A}} \right]\left[ {\rm{B}} \right]}}$
Now, come to the question. Here, the given expression of equilibrium constant is $\dfrac{{{{\left[ {{\rm{N}}{{\rm{H}}_{\rm{3}}}} \right]}^4}{{\left[ {{{\rm{O}}_{\rm{2}}}} \right]}^5}}}{{{{\left[ {{\rm{NO}}} \right]}^4}{{\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]}^6}}}$.
We know that the numerator in the equilibrium constant expression represents concentration of products and the denominator represents concentration of the reaction. So, the reactants are NO and ${{\rm{H}}_{\rm{2}}}{\rm{O}}$ and products are \[{\rm{N}}{{\rm{H}}_{\rm{3}}}\] and ${{\rm{O}}_{\rm{2}}}$. The powers in the molar concentration of reactants or products indicate the stoichiometric coefficient in the balanced chemical equation. So, the balanced chemical equation is,
$4NO+6H_{2}O\rightleftharpoons 4NH_{3}+5O_{2}$
Note: If the chemical reaction is in gaseous phase, that is, reactants and products are all gases, then we can also consider the partial pressure of the species involved instead of their molar concentration. In such conditions, the equilibrium constant is denoted by ${K_p}$
.
For the reaction,
$aA+bB\rightleftharpoons cC+dD$
${K_p} = \dfrac{{p{C^c}.p{D^d}}}{{p{A^a}.p{B^b}}}$
Where, $pC$, $pD$, $pA$ and $pB$ are partial pressures of C,D, A and B respectively.
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