Q.58 Given below are two statements:
Statement–I: Enzymes lower the activation energy of reactions.
Statement–II: The presence of an enzyme has no effect on Keq.
In the light of the above statements, choose the most appropriate answer from the options given below.
- Both Statement–I and Statement–II are true
- Both Statement–I and Statement–II are false
- Statement–I is true but Statement–II is false
- Statement–I is false but Statement–II is true
Correct Answer: Both Statement–I and Statement–II are true.
Enzymes accelerate reactions by reducing activation energy while leaving equilibrium unchanged. Both statements reflect fundamental enzyme kinetics principles.
Statement–I Analysis
Enzymes lower activation energy (Ea) by stabilizing the transition state through active site binding, strain induction, or acid-base catalysis. This provides an alternative pathway with lower Ea, increasing reaction rate without altering ΔG.
Statement–II Analysis
Enzymes do not affect Keq (equilibrium constant), which depends solely on ΔG° = -RT ln Keq. Enzymes speed attainment of equilibrium but cannot shift reactant ⇌ product ratios, as they catalyze both forward and reverse reactions equally.
Option Breakdown
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Both true: Correct; Statement–I describes kinetic acceleration, Statement–II thermodynamic neutrality—core enzyme properties.
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Both false: Incorrect; both are textbook biochemistry facts.
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Statement–I true, Statement–II false: Wrong; Keq invariance is fundamental.
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Statement–I false, Statement–II true: Invalid; Ea reduction defines catalysis.
Introduction to Enzymes Lower Activation Energy Keq
Enzymes lower activation energy of reactions while maintaining no effect on Keq, enabling rapid equilibrium attainment without position shift. This dual property defines biological catalysis efficiency.
Activation Energy Reduction Mechanism
Enzymes stabilize transition states via proximity/orientation effects, electrostatic stabilization, or covalent catalysis. Ea drops from ~50-100 kJ/mol to 20-40 kJ/mol, yielding 10³-10⁶ rate enhancements.
Equilibrium Constant Invariance
Keq =
/[reactants] at equilibrium remains unchanged since enzymes accelerate both directions proportionally. Example: Hexokinase speeds glucose + ATP → G6P + ADP (Vf) and reverse (Vr) equally, preserving Keq ~ 200.$39.00
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Exam Context for Biochemistry
GATE Life Sciences tests this via Michaelis-Menten vs. thermodynamic concepts. Inhibitors affect Vmax/Km but not Keq; uncouplers collapse gradients without equilibrium shift—key for your molecular biology focus.
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