
In a reversible chemical reaction at equilibrium, if the concentration of any one of the reactants is double, then the equilibrium constant will:
A. Also be doubled
B. Be halved
C. Remains the same
D. Becomes one-fourth
Answer
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Hint: To solve this question, first we need to understand the concept of equilibrium constant. We can define equilibrium constant as the value of a chemical reaction quotient when the reaction reaches the state of equilibrium. The value of equilibrium constant is independent of the concentration of a reactant and product. Also, it depends on the temperature and on ionic strength of the mixture.
Complete step by step answer:As we know that, the product of the molar concentration of the products raised to power equal to its stoichiometric coefficient and divided by the product of the molar concentrations of the reactants raised to the power of its stoichiometric coefficient at constant temperature termed as the equilibrium constant of a reaction.
From this, we can understand that the equilibrium constants cannot be changed when the change in concentration of a reaction occurs. It changes only when there is a change of temperature.
Also, the position of equilibrium constant can be changed if the concentration of things present in the mixture changes. With reference to Le Chatelier's Principle, we can say that the position of equilibrium moves in such a way that it tends to undo the change that occurred in the reaction.
Therefore, we can state that if we increase the concentration of the reactant, then the position of equilibrium will move towards right which decreases the concentration of reactant.
So, according to the question, in a reversible chemical reaction at equilibrium, if the concentration of any one of the reactants is double, then the equilibrium constant will remain the same (option C) throughout the reaction.
Therefore, the option C is correct.
Note:For solving this question, we need to keep in mind the laws of chemical equilibrium, we can define that the direction of a chemical reaction will proceed, when the quantities of reactants and products will remain the same after the reaction ends.
Complete step by step answer:As we know that, the product of the molar concentration of the products raised to power equal to its stoichiometric coefficient and divided by the product of the molar concentrations of the reactants raised to the power of its stoichiometric coefficient at constant temperature termed as the equilibrium constant of a reaction.
From this, we can understand that the equilibrium constants cannot be changed when the change in concentration of a reaction occurs. It changes only when there is a change of temperature.
Also, the position of equilibrium constant can be changed if the concentration of things present in the mixture changes. With reference to Le Chatelier's Principle, we can say that the position of equilibrium moves in such a way that it tends to undo the change that occurred in the reaction.
Therefore, we can state that if we increase the concentration of the reactant, then the position of equilibrium will move towards right which decreases the concentration of reactant.
So, according to the question, in a reversible chemical reaction at equilibrium, if the concentration of any one of the reactants is double, then the equilibrium constant will remain the same (option C) throughout the reaction.
Therefore, the option C is correct.
Note:For solving this question, we need to keep in mind the laws of chemical equilibrium, we can define that the direction of a chemical reaction will proceed, when the quantities of reactants and products will remain the same after the reaction ends.
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