Question:
An enzyme-catalyzed reaction is found to have \(\Delta G = -100\) kJ mol\(^{-1}\). Which of the following statements about this reaction is/are true?
An enzyme-catalyzed reaction is found to have \(\Delta G = -100\) kJ mol\(^{-1}\). Which of the following statements about this reaction is/are true?
Show Hint
\(\Delta G\) fixes only the equilibrium position (via \(K\)), not the reaction rate; a very large negative \(\Delta G\) gives a huge \(K\), making the reaction effectively one-directional.
Updated On: Aug 7, 2026
- The rate of the reaction cannot be predicted
- The rate of the reaction is high
- The rate of the reaction is low
- The reaction is irreversible
Show Solution
Verified By Collegedunia
The Correct Option is A, D
Solution and Explanation
Step 1: Separate thermodynamics from kinetics.
The Gibbs free energy change (\(\Delta G\)) of a reaction tells us whether the reaction is thermodynamically favorable and how far it can proceed toward products at equilibrium. It says nothing about how fast the reaction proceeds.
The rate of a reaction depends on the activation energy barrier and the rate constant, which come from the reaction kinetics, not from \(\Delta G\) alone.
So even a reaction with a very large negative \(\Delta G\) can be slow if its activation energy is high; an enzyme only lowers this barrier, it does not change \(\Delta G\).
Step 2: Check option A against this idea.
Because \(\Delta G = -100\) kJ mol\(^{-1}\) only fixes the equilibrium position, the actual rate could be fast or slow depending on the enzyme's turnover number and the activation energy of the catalyzed path.
So statement A, "the rate of the reaction cannot be predicted", is correct.
Step 3: Rule out options B and C.
Option B says the rate is high and option C says the rate is low. Both try to fix a rate value from \(\Delta G\) alone, which step 1 shows is not possible.
Since \(\Delta G\) and rate are independent quantities, both B and C are wrong for the same reason.
Step 4: Use \(\Delta G\) to test reversibility (option D).
At equilibrium, \(\Delta G^{\circ} = -RT \ln K\), so
\[ \ln K = \frac{-\Delta G}{RT} = \frac{100000\ \text{J mol}^{-1}}{8.314\ \text{J mol}^{-1}\text{K}^{-1} \times 298\ \text{K}} \approx 40.4 \]
This gives \(K \approx e^{40.4} \approx 3.5 \times 10^{17}\), an enormous equilibrium constant.
Such a huge \(K\) means the equilibrium lies almost completely on the product side, so the reverse reaction is negligible and the reaction behaves as irreversible in practice.
So statement D is correct.
Final Answer:
The rate cannot be predicted from \(\Delta G\) alone, and the very large negative \(\Delta G\) makes the reaction effectively irreversible.
\[ \boxed{\text{Options A and D}} \]
The Gibbs free energy change (\(\Delta G\)) of a reaction tells us whether the reaction is thermodynamically favorable and how far it can proceed toward products at equilibrium. It says nothing about how fast the reaction proceeds.
The rate of a reaction depends on the activation energy barrier and the rate constant, which come from the reaction kinetics, not from \(\Delta G\) alone.
So even a reaction with a very large negative \(\Delta G\) can be slow if its activation energy is high; an enzyme only lowers this barrier, it does not change \(\Delta G\).
Step 2: Check option A against this idea.
Because \(\Delta G = -100\) kJ mol\(^{-1}\) only fixes the equilibrium position, the actual rate could be fast or slow depending on the enzyme's turnover number and the activation energy of the catalyzed path.
So statement A, "the rate of the reaction cannot be predicted", is correct.
Step 3: Rule out options B and C.
Option B says the rate is high and option C says the rate is low. Both try to fix a rate value from \(\Delta G\) alone, which step 1 shows is not possible.
Since \(\Delta G\) and rate are independent quantities, both B and C are wrong for the same reason.
Step 4: Use \(\Delta G\) to test reversibility (option D).
At equilibrium, \(\Delta G^{\circ} = -RT \ln K\), so
\[ \ln K = \frac{-\Delta G}{RT} = \frac{100000\ \text{J mol}^{-1}}{8.314\ \text{J mol}^{-1}\text{K}^{-1} \times 298\ \text{K}} \approx 40.4 \]
This gives \(K \approx e^{40.4} \approx 3.5 \times 10^{17}\), an enormous equilibrium constant.
Such a huge \(K\) means the equilibrium lies almost completely on the product side, so the reverse reaction is negligible and the reaction behaves as irreversible in practice.
So statement D is correct.
Final Answer:
The rate cannot be predicted from \(\Delta G\) alone, and the very large negative \(\Delta G\) makes the reaction effectively irreversible.
\[ \boxed{\text{Options A and D}} \]
Was this answer helpful?
0
0
Top GATE BT Biochemistry Questions
- An enzyme (E) catalyzes the biochemical reaction \(A \rightarrow B\) with \(k_{\text{cat}}=500~\text{s}^{-1}\). If the initial reaction velocity \((V_0)\) is \(10~\mu\text{M}\cdot\text{s}^{-1}\) at the total enzyme concentration \([E_t]=30~\text{nM}\) and substrate concentration \([A]=40~\mu\text{M}\), the value of \(K_m\) (in \(\mu\text{M}\)) is ________________.
- Intracellular proteins are targeted for proteolytic degradation in proteasomes upon conjugation with
- The number of different possible ways of forming five intramolecular disulfide bonds with ten cysteine residues of a protein is ________________.
- Which of the following enzyme is involved in the protection of cells from reactive oxygen species?
- You have purified an enzyme using a series of chromatographic methods. It was observed that a 10 \(\mu\)g mL-1 of this purified enzyme converted 10 mM substrate per hour at 25\(^{\circ}\)C and pH 7. Its specific activity is IU \(\mu\)g-1. (rounded off to three decimal places)
Top GATE BT Questions
- In adsorption chromatography, the adsorption of uncharged solute molecules onto a silica-based stationary phase is by
- The transfer function of a process is \( G(s) = \frac{K_p}{T_p s + 1} \), where \( K_p \) is the gain and \( T_p \) is the time constant. This is a _____ process.
- Which one of the following statements is correct in the context of thermodynamics?
- Which one of the following statements is correct about Reynolds Number (NRe) in a stirred tank bioreactor?
- The relationship that involves the exchange of nutrients between two different species for their mutual growth is called___
View More Questions