NEET UG CHEMISTRY MCQ: THERMODYNAMICS

Solution approach

1. Correct Answer + Core Reason: $\Delta \,{G^ \circ } < 0,\,K < 1\,$. When $\Delta G^\circ < 0$, K > 1.
   
   Step-by-Step Solution:
   
   Step 1: Recall the relationship between Gibbs free energy and the equilibrium constant: $\Delta G^\circ = -RT\ln K$
   
   Step 2: Analyze the relationship:
   - If $\Delta G^\circ < 0$, then $-RT\ln K < 0$, which means $\ln K > 0$, and therefore $K > 1$.
   - If $\Delta G^\circ = 0$, then $-RT\ln K = 0$, which means $\ln K = 0$, and therefore $K = 1$.
   - If $\Delta G^\circ > 0$, then $-RT\ln K > 0$, which means $\ln K < 0$, and therefore $K < 1$.
   
   Step 3: Compare the derived relationships with the given options. Option D is incorrect because it states that when $\Delta G^\circ < 0$, $K < 1$, which is the opposite of what is expected.
   
   Why other options are wrong:
   - **Option A ($\Delta \,{G^ \circ } < 0,\,K > 1$):** This is correct because a negative $\Delta G^\circ$ indicates a spontaneous reaction, which favors product formation, leading to $K > 1$.
   - **Option B ($\Delta \,{G^ \circ } = 0,\,K = 1$):** This is correct because when $\Delta G^\circ$ is zero, the reaction is at equilibrium, and $K = 1$.
   - **Option C ($\Delta \,{G^ \circ } > 0,\,K < 1$):** This is correct because a positive $\Delta G^\circ$ indicates a non-spontaneous reaction, which favors reactant formation, leading to $K < 1$.
   
   Key Concepts:
   - Gibbs free energy and equilibrium constant: $\Delta G^\circ = -RT\ln K$
   - Spontaneity and equilibrium.
   
   Memory Aids:
   - Remember the relationship between Gibbs free energy and the equilibrium constant.
   
   Quick Tips:
   - Understand the conditions under which each equation is valid.
   - Pay attention to the signs.