In: Chemistry
The reaction 1,3-bisphosphoglycerate + ADP <=> 3-phosphoglycerate + ATP has ΔG° = -18.8 kJ mol-1. Calculate ΔGrxn in kJ mol-1 at 37.0 °C when [1,3-bisphosphoglycerate] = 1.90 mmol L-1, [ADP] = 1.20 mmol L-1, [3-phosphoglycerate] = 0.450 mmol L-1, and [ATP] = 4.00 mmol L-1. (R = 8.3145 J mol-1 K-1)
The reaction S(aq) + T(aq) <=> U(aq) has K = 17.5. Which of the following is true when [S] = 12.0 mmol L-1, [T] = 19.0 mmol L-1, and [U] = 250 mmol L-1? The reaction is forming products There is insufficient information to answer this question The reaction is in equilibrium The reaction is forming reactants
Calculate ΔGrxn for a reaction (in kJ mol-1) that has ΔG° = -4.70 kJ mol-1 and Q = 850 at 30 °C. (R = 8.3145 J mol-1 K-1)
1) The reaction 1,3-bisphosphoglycerate + ADP <=> 3-phosphoglycerate + ATP has ΔG° = -18.8 kJ mol-1. Calculate ΔGrxn in kJ mol-1 at 37.0 °C when [1,3-bisphosphoglycerate] = 1.90 mmol L-1, [ADP] = 1.20 mmol L-1, [3-phosphoglycerate] = 0.450 mmol L-1, and [ATP] = 4.00 mmol L-1. (R = 8.3145 J mol-1 K-1)
Answer :
Given: ΔG° = -18.8 kJ mol-1. T = 37.0 °C = 37 + 273 = 310 K
and [1,3-bisphosphoglycerate] = 1.90 mmol L-1, [ADP] = 1.20 mmol L-1,
[3-phosphoglycerate] = 0.450 mmol L-1, and [ATP] = 4.00 mmol L-1.
R = 8.3145 J mol-1 K-1 = 0.0083145 J mol-1 K-1)
Balanced phisiological conversion :
1,3-bisphosphoglycerate + ADP <--------> 3-phosphoglycerate + ATP
Q = [3-phosphoglycerate][ATP] / [1,3-bisphosphoglycerate][ADP]
Using given concentrations,
Q = (4.00)(0.45) / (1.90)(1.20)
Q = 0.789
Gibbs equation giving Free energy change is
GRXN = Go + RT ln Q
Using given data :
GRXN = (-18.8) + (0.0083145)(310) ln (0.789)
GRXN = -18.8 + (-0.6)
GRXN = -19.4 kJ.mole-1.
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2) The balanced reaction :
S(aq) + T(aq) <--------> U(aq)
Given : ΔG° = -4.70 kJ mol-1 and Q = 850 , T = 30 °C = 30 + 273 = 303 K
R = 0.0083145 J mol-1 K-1.
Using equation,
GRXN = Go + RT ln Q
GRXN = (-4.70) + 0.0083145 x 303 x ln(850)
GRXN = (-4.70) + (+6.75)
GRXN = +2.05 kJ.mole-1.
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