For a reversible reaction: A + B ⇌ C, if the concentrations of the reactants are doubled at a definite temperature, then equilibrium constant will

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Prashant Kotian · Accepted Answer

Correct answer: D. Step 1 - Express the Equilibrium Constant ($K_c$) For the given general reversible reaction: $$\ce{A + B <=> C}$$ The equilibrium constant in terms of molar concentrations ($K_c$) is expressed by the law of chemical equilibrium as the ratio of the equilibrium concentration of the product to the product of the equilibrium concentrations of the reactants: $$K_c = \frac{[\ce{C}]}{[\ce{A}][\ce{B}]}$$ Step 2 - Understand the Thermodynamic Basis of the Equilibrium Constant According to chemical thermodynamics, the standard Gibbs free energy change of a reaction ($\Delta G^\circ$) is related to the equilibrium constant by the equation: $$\Delta G^\circ = -RT \ln K_c$$ Where: * $\Delta G^\circ$ is the standard Gibbs free energy change of the reaction. * $R$ is the universal gas constant. * $T$ is the absolute temperature in Kelvin ($\text{K}$). Since $\Delta G^\circ$ is a constant for a given reaction at a specific temperature, the equilibrium constant $K_c$ depends strictly on temperature only: $$K_c = f(T)$$ Step 3 - Analyze the Effect of Doubling Reactant Concentrations At a definite (constant) temperature, if we double the initial concentrations of the reactants $\ce{A}$ and $\ce{B}$, the reaction quotient ($Q_c$) momentarily decreases: $$Q_c = \frac{[\ce{C}]}{[2\ce{A}][2\ce{B}]} = \frac{1}{4} K_c < K_c$$ Because the reaction quotient $Q_c$ is less than $K_c$, the system is no longer at equilibrium. According to Le Chatelier's principle: * The system will spontaneously shift in the forward direction ($\ce{A + B -> C}$) to counteract this change. * Reactants $\ce{A}$ and $\ce{B}$ will be consumed, and more of product $\ce{C}$ will be formed. * This forward shift continues until the concentrations adjust such that the reaction quotient $Q_c$ once again becomes equal to the value of $K_c$. Therefore, while changing the initial concentrations alters the individual equilibrium concentrations of the species, the ratio of product concentration to reactant concentration at the new equilibrium remains constant. The value of the equilibrium constant $K_c$ is unchanged. $$K_c = \boxed{\text{remain same}}$$ Step 4 - Evaluate the Options * **Option (A) "be doubled"**: Incorrect. The equilibrium constant is not directly proportional to the reactant concentrations. * **Option (B) "be halved"**: Incorrect. * **Option (C) "be one fourth"**: Incorrect. This is a common distractor representing the temporary, instantaneous value of the reaction quotient $Q_c$ immediately after concentrations are doubled, before the reaction shifts to re-establish equilibrium. * **Option (D) "remain same"**: Correct. Since the temperature is held constant, the equilibrium constant remains entirely unchanged.

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