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The reaction 1,3-bisphosphoglycerate + ADP <=> 3-phosphoglycerate + ATP has ΔG° = -18.8 kJ mol-1. Calculate...

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)

Solutions

Expert Solution

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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In Q.2 ) Middle some information is not clear enough.


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