
he equilibrium constant ${K_c}$ for the reaction will be:
(A) antilog 15.6
(B) antilog 2.5
(C) antilog 1.5
(D) antilog 12.2
Answer
221.4k+ views
Hint - To solve this question we will define and then calculate the equilibrium constant ${K_c}$, which is the ratio of the equilibrium concentrations of products over the equilibrium concentration of reactants each raised to the value of their stoichiometric coefficients.
Complete answer:
> In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations having no further tendency to change with time, so that there is no change in the properties of the system, such a state is known as equilibrium.
> The equilibrium constant of a reaction (donated as ${K_c}$) provides an insight into the relationship between the reactants and products when the reaction is at equilibrium. There are different types of equilibrium constants which state the relationship between the reactants and products present in equilibrium reactions in the terms of different rules.
> Equilibrium constant has no units as active masses are used instead of molarity/concentrations of the respective substances. And as we know, active mass is a unitless ratio of mass reaction to the total mass present..
At equilibrium,
- Rate (forward reaction) = Rate (backward reaction), i.e.
$ \Rightarrow \,\,{r_f}\, = \,\,\,{r_b}$
Now let’s come to the solution for this question-
The equilibrium constant${K_c}$ for the reaction -
$E_{cell}^ \circ = E_{reduction}^ \circ - E_{Oxidation}^ \circ $
${E^ \circ } = \dfrac{{0.059}}{n}\log {K_c}$
$0.46 = \dfrac{{0.059}}{2}\log {K_c}$
$\log {K_c} = \dfrac{{0.46 \times 2}}{{0.059}} = 15.6$
${K_c} = anti\log 15.6$
Hence, the answer is option A. antilog 15.6
Note - Equilibrium constant can be defined as the ratio of the equilibrium concentrations of products over the equilibrium concentration of reactants each raised to the power of their stoichiometric coefficients. There are different types of equilibrium constants which state the relationship between products and reactants present in equilibrium in terms of different rules.
Complete answer:
> In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations having no further tendency to change with time, so that there is no change in the properties of the system, such a state is known as equilibrium.
> The equilibrium constant of a reaction (donated as ${K_c}$) provides an insight into the relationship between the reactants and products when the reaction is at equilibrium. There are different types of equilibrium constants which state the relationship between the reactants and products present in equilibrium reactions in the terms of different rules.
> Equilibrium constant has no units as active masses are used instead of molarity/concentrations of the respective substances. And as we know, active mass is a unitless ratio of mass reaction to the total mass present..
At equilibrium,
- Rate (forward reaction) = Rate (backward reaction), i.e.
$ \Rightarrow \,\,{r_f}\, = \,\,\,{r_b}$
Now let’s come to the solution for this question-
The equilibrium constant${K_c}$ for the reaction -
$E_{cell}^ \circ = E_{reduction}^ \circ - E_{Oxidation}^ \circ $
${E^ \circ } = \dfrac{{0.059}}{n}\log {K_c}$
$0.46 = \dfrac{{0.059}}{2}\log {K_c}$
$\log {K_c} = \dfrac{{0.46 \times 2}}{{0.059}} = 15.6$
${K_c} = anti\log 15.6$
Hence, the answer is option A. antilog 15.6
Note - Equilibrium constant can be defined as the ratio of the equilibrium concentrations of products over the equilibrium concentration of reactants each raised to the power of their stoichiometric coefficients. There are different types of equilibrium constants which state the relationship between products and reactants present in equilibrium in terms of different rules.
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