Q.53 The concentrations of ATP, ADP and inorganic phosphate in a cell are 2.59, 0.73 and 2.72 mM, respectively. Under these conditions, free energy change for the synthesis of ATP at 37 °C is. _ kJ/mol (round off to 2 decimal places) Given: free energy change for ATP hydrolysis under standard conditions is -30.5 kJ/mol and R = 8.315 kJ/mol.K

Q.53 The concentrations of ATP, ADP and inorganic phosphate in a cell are 2.59, 0.73
and 2.72 mM, respectively. Under these conditions, free energy change for the
synthesis of ATP at 37 °C is. _ kJ/mol
(round off to 2 decimal places)
Given: free energy change for ATP hydrolysis under standard conditions is 30.5 kJ/mol and R = 8.315 kJ/mol.K

Free Energy Change for ATP Synthesis at 37°C: 2.59 mM ATP, 0.73 mM ADP, 2.72 mM Pi Calculation

The free energy change for ATP synthesis under the given cellular conditions is +35.81 kJ/mol. This value is calculated using the standard equation for non-standard Gibbs free energy, reflecting the endergonic nature of ATP formation.

🧮 Calculation Steps

The reaction for ATP synthesis is the reverse of hydrolysis: ADP + Pi → ATP + H2O

ΔG = ΔG°’ + RT ln(Q)
where ΔG°’ for synthesis = +30.5 kJ/mol (negative of hydrolysis)
R = 8.315 × 10-3 kJ/mol·K, T = 310 K (37°C)
Q = [ATP] / ([ADP][Pi])

Step 1: Convert concentrations to M

  • [ATP] = 0.00259 M
  • [ADP] = 0.00073 M
  • [Pi] = 0.00272 M

Step 2: Calculate Reaction Quotient (Q)

Q = 0.00259 / (0.00073 × 0.00272) = 1302.14

Step 3: Calculate RT ln(Q)

RT = (8.315 × 10-3) × 310 = 2.577 kJ/mol
ln(1302.14) ≈ 7.174
RT ln(Q) = 2.577 × 7.174 = 18.31 kJ/mol

Step 4: Final ΔG Calculation

ΔG = 30.5 + 18.31 = +35.81 kJ/mol

📋 Detailed Breakdown

  • ΔG°’: +30.5 kJ/mol for synthesis (hydrolysis releases -30.5 kJ/mol under standard conditions: 1 M, pH 7)
  • Temperature: 37°C = 310 K (standard physiological temperature)
  • Q Ratio: High [ATP]/[ADP][Pi] = 1302 makes synthesis unfavorable (ΔG > 0)
  • Result: Rounded to two decimal places: 35.81 kJ/mol

🎯 Why ΔG is Positive Here

The mass-action ratio Q = 1302 far exceeds the equilibrium constant, making ΔG = +35.81 kJ/mol. This positive value indicates ATP synthesis requires energy input, unlike hydrolysis which is exergonic.

🏆 Step-by-Step for Exam Success (IIT JAM/GATE)

  1. Convert T to Kelvin: 37°C = 310 K
  2. Calculate Q = [ATP]/([ADP][Pi]) = 1302
  3. RT ln(Q) = 2.577 kJ/mol × ln(1302) ≈ +18.31 kJ/mol
  4. ΔG = +30.5 + 18.31 = +35.81 kJ/mol

💡 Exam Tip:

This matches GATE BT 2020 Q.53 pattern. Always check units (mM → M, kJ consistency) and remember ΔG°’ for synthesis is +30.5 kJ/mol.

 

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