Mass Action Ratio

Tutorial Question 1: Mass Action Ratio Consider the reaction: A+B ↔C +D At a given point, the concentrations of the reactants and products are as follows: • [A] = 1.0 M • [B] = 2.0 M • [C] = 0.5 M • [D] = 0.5 M The equilibrium constant for this reaction is Keq = 1.0. a. Calculate the mass action ratio Q. b. Determine the direction of the reaction. Solution: a. The mass action ratio Q is given by: Q= [C][D] [A][B] Substituting the given values: Q= 0.25 (0.5)(0.5) = = 0.125 (1.0)(2.0) 2.0 b. Since Q = 0.125 and Keq = 1.0, we have Q < Keq . This indicates that the reaction is not at equilibrium, and it will proceed in the forward direction to produce more C and D until Q equals Keq . Question 2: Membrane Potential of Uncharged Molecules A cell membrane separates two compartments with different concentrations of glucose: • [Glucose]outside = 10 mM • [Glucose]inside = 1 mM The temperature is 37◦ C (310 K). a. Derive and calculate the free energy change (∆G) for the transport of glucose from the outside to the inside of the cell. Use R = 8.314 J mol−1 K−1 . b. Explain the significance of the calculated ∆G value in terms of energy requirements for glucose transport. Solution: a. The free energy change for uncharged molecules is given by:   [Glucose]inside ∆G = RT ln [Glucose]outside Substituting the given values:  ∆G = (8.314)(310) ln 1 10  Simplify: ∆G = 2577.34 ln(0.1) ≈ 2577.34(−2.302) ∆G ≈ −5932 J/mol ≈ −5.93 kJ/mol b. A negative ∆G indicates that the transport of glucose from the outside to the inside of the cell is energetically favorable and does not require the input of energy (i.e., it is passive transport). 1 Question 3: Water and Osmosis A cell is placed in a solution with the following conditions: • Intracellular solute concentration: 300 mM • Extracellular solute concentration: 200 mM a. Predict the direction of water movement. b. Describe the effect on the cell volume over time. c. Explain the concept of osmosis and its relationship to solute concentration. Solution: a. Water will move into the cell. b. As water enters the cell, the cell volume will increase. If too much water enters, the cell may burst (lyse). c. Osmosis is the movement of water across a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration. This occurs to equalize solute concentrations on both sides of the membrane. Question 4: Ion Transport and Membrane Potential A neuron has the following ionic concentrations for K+ : • [K+ ]inside = 140 mM • [K+ ]outside = 4 mM The membrane potential is −70 mV, and the temperature is 37◦ C (310 K). Use the Nernst equation to calculate the equilibrium potential for K+ :  +  [K ]outside RT ln Eion = zF [K+ ]inside where z = +1, F = 96, 500 C/mol. Solution: a. Substitute the given values into the Nernst equation: Eion (8.314)(310) = ln (1)(96, 500)  4 140  Simplify: Eion = 0.0267 ln(0.0286) ≈ 0.0267(−3.555) Eion ≈ −0.095V = −95 mV b. The equilibrium potential (Eion ) represents the voltage at which there is no net movement of K+ across the membrane. Since the actual membrane potential (−70 mV) differs from Eion , there will be a driving force for K+ ions to move. Question 5: Proton Motive Force In chloroplasts, the ∆pH across the thylakoid membrane is 3.5. Calculate the proton motive force (pmf) using the formula: RT ∆pH pmf = 2.303 F Use R = 8.314 J mol−1 K−1 , T = 298 K, F = 96, 500 C/mol. Solution: 2 a. Substitute the given values into the formula: pmf = 2.303 · (8.314)(298) · 3.5 96, 500 Simplify: pmf = 2.303 · 0.0256 · 3.5 ≈ 0.208 V b. The proton motive force is a measure of the energy available to drive protons across the membrane. This energy can be used for ATP synthesis or other cellular processes. 3