I can't seem to get this problem, i really don't understand it
Chemical reactions can also be coupled. The decreasing free energy of a spontaneous change can be stored in substance whose formation would not ordinarily be spontaneous. Living organisms have a remarkable ability to couple reactions in order to make the most of the free energy that passes through. The principle biological free- energy reservoir is the ATP-ADP system. When adenosine diphosphate (ADP) molecules add phosphate units to produce adenosine triphosphate( ATP) molecules, they store 31 kj/mol of free energy that can be release when the reaction is reversed :
ADP^(3-)(aq)+ HPO4^(2-)(aq)+H^(+) „³ ATP^(4-)(aq) +H2O(l) delta G =31kJ
In a living cell, this free energy is supplied by the oxidation of glucose to CO2 and H2O:
C6H12O6(s)+6O2(g)„³6CO2(g)+6H2O(l) delta G = -2870kJ
The oxidationof 1 mol of glucose releses 2870 kJ of free energy. Some of this energy is used to convert 36 mol of ADP to ATP:
36ADP^3-(aq) + 36HPO4^2-(aq) +36H^+(aq) -->36ATP^4(aq)- + 36H2O(l) delta G = (36)(31)= 1100kJ
Hence, 1100kJ, or 38%, of the energy released by 1 mol of glucose is stored in ATP. The ATP later releases this energy as needed. Some of the energy is used for muscle action, and some is used to synthesize highly organized biomolecules and structural elements. In this way, living organisms are able to exist as temporary islands of low entropy in an increasingly disordered environment. Eventually, the stored energy is released, and the universe obtains its full entropy increase.
Questions:
1)Calculate the equilibrium constant at 25 degrees Celsius for the hydrolosis of adenosine triphosphate (ATP) into adenosine diphosphate (ADP):
ATP^(4-)+ H2O(l)<-„³ADP^(3-)(aq)+HPO4^(2-)(aq)+H^(+)(aq)
2)In human red blood cells, the concentrations of ATP^(4-), ADP^(3-) and HPO4^(2-) are 2.25 mM, 0.25mM and 1.65 mM respectively. Estimate delta G and use it to show which reaction, forward or reverse, is capable of providing useful work.
Assume that the pH is 7 and the temperature is 25 degrees Celsius .
To solve these problems, we need to use the relationship between the equilibrium constant (K), free energy change (ΔG), and concentration of reactants and products. The equation is given by:
ΔG = -RTlnK
where ΔG is the free energy change, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and ln is the natural logarithm.
Let's solve the problems step by step:
1) Calculate the equilibrium constant (K) at 25 degrees Celsius for the hydrolysis of adenosine triphosphate (ATP) into adenosine diphosphate (ADP):
The equation for the hydrolysis reaction is:
ATP^(4-) + H2O(l) ←→ ADP^(3-) + HPO4^(2-) + H^(+)
Since the reaction is at equilibrium, the forward and reverse reactions are balanced. Therefore, we can assume that the concentrations of the species on both sides of the equation will be equal at equilibrium.
We can use the concentrations of the products and reactants given in the problem to calculate the equilibrium constant. However, we need to convert the concentrations from millimoles per liter (mM) to moles per liter (M). We can do this by dividing the concentrations by 1000.
Given concentrations:
[ATP^(4-)] = 2.25 mM = 2.25/1000 M
[ADP^(3-)] = 0.25 mM = 0.25/1000 M
[HPO4^(2-)] = 1.65 mM = 1.65/1000 M
Using these concentrations, we can substitute them into the equilibrium constant expression:
K = [ADP^(3-)] × [HPO4^(2-)] × [H^(+)] / [ATP^(4-)] × [H2O(l)]
Since the concentration of water is relatively constant and does not affect the equilibrium constant significantly, we can assume it is constant and cancel it out:
K = [ADP^(3-)] × [HPO4^(2-)] × [H^(+)] / [ATP^(4-)]
Substituting the given concentrations:
K = (0.25/1000) × (1.65/1000) × 1 / (2.25/1000)
Simplifying the expression:
K ≈ 0.003629
Therefore, the equilibrium constant for the hydrolysis of ATP into ADP is approximately 0.003629 at 25 degrees Celsius.
2) Estimate ΔG and use it to show which reaction, forward or reverse, is capable of providing useful work:
To estimate ΔG, we can use the formula:
ΔG = ΔG° + RTlnQ
where ΔG° is the standard free energy change, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, ln is the natural logarithm, and Q is the reaction quotient.
Given:
[ATP^(4-)] = 2.25 mM = 2.25/1000 M
[ADP^(3-)] = 0.25 mM = 0.25/1000 M
[HPO4^(2-)] = 1.65 mM = 1.65/1000 M
To calculate the reaction quotient (Q), we need to substitute the concentrations of the reactants and products:
Q = [ADP^(3-)] × [HPO4^(2-)] × [H^(+)] / [ATP^(4-)] × [H2O(l)]
Since the concentration of water is relatively constant and does not significantly affect the reaction quotient, we can assume it is constant and cancel it out:
Q = [ADP^(3-)] × [HPO4^(2-)] × [H^(+)] / [ATP^(4-)]
Substituting the given concentrations:
Q = (0.25/1000) × (1.65/1000) × 1 / (2.25/1000)
Simplifying the expression:
Q ≈ 0.003629
Now, we can substitute the values into the equation for ΔG:
ΔG = ΔG° + RTlnQ
Given:
ΔG° = 31 kJ/mol = 31 × 1000 J/mol
T = 25 + 273.15 K
Substituting the values:
ΔG = (31 × 1000) + (8.314 × (25 + 273.15)) × ln(0.003629)
Calculating the value of ΔG using a calculator:
ΔG ≈ -18,990 J/mol ≈ -19 kJ/mol
Since ΔG is negative, the reaction is exergonic and favors the formation of ATP from ADP. Therefore, the reverse reaction, ATP hydrolysis into ADP, is capable of providing useful work.
To summarize, the equilibrium constant for the hydrolysis of ATP into ADP at 25 degrees Celsius is approximately 0.003629. Additionally, the ΔG value for the reaction indicates that the reverse reaction, ATP hydrolysis into ADP, is capable of providing useful work.