
The equilibrium constants for the reaction, $ B{r_2}\overset {} \leftrightarrows 2Br $ at $ 500K $ and $ 700K $ are $ 1 \times {10^{ - 10}} $ and $ 1 \times {10^{ - 5}} $ respectively. The reaction is:
(A) Endothermic
(B) Exothermic
(C) Fast
(D) Slow
Answer
497.7k+ views
Hint: The equilibrium constant for the chemical reaction is the reaction quotient at the chemical equilibrium. It is represented by $ {K_c} $ . The equilibrium constant of a chemical reaction is expressed as the ratio of the concentration of products to the concentration of the reactants. The equilibrium constant depends upon the temperature and several other factors.
Formula used:
For the chemical reaction, $ A + B\overset {} \leftrightarrows xC $ then the equilibrium concentration $ {K_c} $ for the reaction will be, $ {K_c} = \dfrac{{{{\left[ C \right]}^x}}}{{\left[ A \right]\left[ B \right]}} $ where, $ [A],[B] $ and $ [C] $ are the concentrations of reactant and products respectively.
Complete step by step answer:
Now we know the basic concept of chemical equilibrium and the equilibrium constant of the reaction. Now we will use the basic concept of chemical equilibrium to predict the nature of chemical reactions. So let’s start with the question and consider the given reaction $ B{r_2}\overset {} \leftrightarrows 2Br $ . In the question, we have given the equilibrium constants at different temperatures. The given values are $ {K_c} = 1 \times {10^{ - 10}} $ at $ T = 500K $ and $ {K_c} = 1 \times {10^{ - 5}} $ at $ T = 700K $ . From the given quantities we can observe that the equilibrium constant depends on the temperature. The equilibrium constant increases with the increase in temperature. So, now for the reaction, $ B{r_2}\overset {} \leftrightarrows 2Br $ the equilibrium constant is given $ {K_c} = \dfrac{{{{\left[ {Br} \right]}^2}}}{{[B{r_2}]}} $ . So the equilibrium constant of the reaction $ {K_c} $ increases when the concentration of the product increases and we know that the increase in the concentration of product with an increase in temperature shifts the reaction in the forward direction. We know that the reaction which shifts in the forward direction is endothermic.
Therefore, the correct option is (A).
Note:
When the equilibrium constant of the chemical reaction $ {K_c} $ decreases the reaction shifts in the backward direction and the reaction is considered exothermic. The exothermic reactions are the reaction in which energy is released.
Formula used:
For the chemical reaction, $ A + B\overset {} \leftrightarrows xC $ then the equilibrium concentration $ {K_c} $ for the reaction will be, $ {K_c} = \dfrac{{{{\left[ C \right]}^x}}}{{\left[ A \right]\left[ B \right]}} $ where, $ [A],[B] $ and $ [C] $ are the concentrations of reactant and products respectively.
Complete step by step answer:
Now we know the basic concept of chemical equilibrium and the equilibrium constant of the reaction. Now we will use the basic concept of chemical equilibrium to predict the nature of chemical reactions. So let’s start with the question and consider the given reaction $ B{r_2}\overset {} \leftrightarrows 2Br $ . In the question, we have given the equilibrium constants at different temperatures. The given values are $ {K_c} = 1 \times {10^{ - 10}} $ at $ T = 500K $ and $ {K_c} = 1 \times {10^{ - 5}} $ at $ T = 700K $ . From the given quantities we can observe that the equilibrium constant depends on the temperature. The equilibrium constant increases with the increase in temperature. So, now for the reaction, $ B{r_2}\overset {} \leftrightarrows 2Br $ the equilibrium constant is given $ {K_c} = \dfrac{{{{\left[ {Br} \right]}^2}}}{{[B{r_2}]}} $ . So the equilibrium constant of the reaction $ {K_c} $ increases when the concentration of the product increases and we know that the increase in the concentration of product with an increase in temperature shifts the reaction in the forward direction. We know that the reaction which shifts in the forward direction is endothermic.
Therefore, the correct option is (A).
Note:
When the equilibrium constant of the chemical reaction $ {K_c} $ decreases the reaction shifts in the backward direction and the reaction is considered exothermic. The exothermic reactions are the reaction in which energy is released.
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