Question

Consider the hydrolysis of ATP: ATP(aq) + H2O(l) → ADP(aq) + Pi(aq). This reaction has ΔH°= −24.3 kJ/mol and ΔS°= +21.6 J/mol-K. The actual concentrations of ATP, ADP, and Pi are not 1 M in a biological cell. How much energy can the conversion of ATP to ADP supply when it occurs at physiological conditions in E. coli where the temperature is 37°C and the approximate concentrations are ATP = 11.2 mM, ADP = 1.52 mM, Pi = 20.0 mM? Enter your answer to three significant figures in units of kJ. Include the sign.

Answer 1

Sol.

Reaction :  

ATP(aq) + H2O(l) -----> ADP(aq) + Pi(aq)

As Standard change in enthalpy = deltaH° = - 24.3 KJ / mol

standard change in entropy = deltaS° = 21.6 J / K mol

= 21.6 / 1000 KJ / K mol = 0.0216 KJ / K mol

Temperature = T = 37°C = 37 + 273.15 K = 310.15 K

So , Standard change in free energy

= deltaG° = deltaH° - T × deltaS°

= ( - 24.3 KJ / mol ) - 310.15 K × 0.0216 KJ / K mol

= - 30.99924 KJ / mol

Now , [ADP] = 1.52 mM = 1.52 × 10-3 M

[Pi] = 20 mM = 20 × 10-3 M

[ATP] = 11.2 mM = 11.2 × 10-3 M

Reaction Quotient = Q

= [ADP] [Pi] / [ATP]

= 1.52 × 10-3 × 20 × 10-3 / ( 11.2 × 10-3 )

= 0.0027142857

Gas constant = R = 0.008314 KJ / K mol  

Therefore , Change in free energy

= deltaG = deltaG° + RTln(Q)

= - 30.99924 KJ / mol + 0.008314 KJ / K mol × 310.15 K × ln ( 0.0027142857 )

= - 46.2 KJ / mol  

Therefore ,     - 46.2 KJ    of energy the conversion supply