Thermo - Level 2 Flashcards
(10 cards)
Why can an endothermic reaction still be spontaneous?
Because if the entropy change (ΔS) is positive and the temperature is high enough, the –TΔS term can make ΔG negative.
How does coupling an unfavorable reaction to ATP hydrolysis make it proceed?
The large negative ΔG° from ATP hydrolysis offsets the positive ΔG° of the unfavorable reaction, making the net ΔG° negative and thus spontaneous.
Explain why water freezes spontaneously below 0°C despite a decrease in entropy.
Freezing releases heat (ΔH < 0), and although entropy decreases (ΔS < 0), at low temperatures the –TΔS term is small enough that ΔG is still negative.
If ΔG° = +10 kJ/mol, under what condition can the reaction still be spontaneous?
If Q (reaction quotient) is very small, then RTln(Q) becomes a large negative number and ΔG = ΔG° + RTln(Q) < 0.
Describe a situation where a reaction has ΔG < 0 but does not occur in the body.
If the activation energy is too high, the reaction is kinetically hindered and proceeds too slowly without a catalyst, despite being thermodynamically favorable.
What does it mean when ΔG = 0?
The reaction is at equilibrium—no net change in concentrations of reactants or products.
Why is ΔG more relevant than ΔG° in biological systems?
Because cellular conditions rarely match standard state; ΔG accounts for actual concentrations and determines if a reaction is currently favorable.
Why does ATP have a higher ΔG of hydrolysis in cells compared to standard conditions?
Because the concentrations of ATP, ADP, and Pi are far from 1 M, which makes the RTln(Q) term more negative, decreasing ΔG.
What does a small change in ΔG° mean for the equilibrium constant K?
Because ΔG° = –RTln(K), even small changes in ΔG° cause exponential changes in K due to the logarithmic relationship.
How can you shift a reaction toward products without changing ΔG°?
By decreasing product concentration or increasing reactant concentration, which lowers Q and makes ΔG more negative.
