The formation constants at 25°C for Fe(CN)4-6 and Fe(EDTA)2– are 1.00 x 1037 and 2.10 x 1014, respectively. Answer the questions below. 1) Calculate K under standard conditions for the reaction Fe(EDTA)2−(aq) + 6CN−(aq) ----> Fe(CN)4−6​(aq) + EDTA4−(aq) 2) Calculate ΔG° for the reaction. (kJ/mol)

Given that the AG" values for the hydrolysis of glucose 1-phosphate and glucose 6-phosphate are approximately -21 kJ/mol and –14 kJ/mol, respectively, which statement is true regarding the isomerization shown? glucose 1-phosphate → glucose 6-phosphate O The reaction can likely proceed in either direction and will therefore depend on the concentrations of glucose 1-phosphate and glucose 6-phosphate. O Only the anabolic direction will occur because the free energy change is positive. O The reaction will proceed in both directions simultaneously. This reaction can only proceed in one direction and must be the rate limiting reaction for this point in metabolism.

For the reversible, liquid-phase reaction A ↔ B, determine the adiabatic equilibrium temperature and conversion when pure A is fed to the reactor at 300 K. CPA = CPB = 50 cal mol‍¹ K-¹. Heat capacities are independent of temperature. The equilibrium constant is 1.0 × 105 at 298 K; the heat of reaction is -2.0 × 10 cal/mol.

The reaction quotient is Q=1.6×10-26 Part B What pH is needed to produce this value of Q if the concentration and pressure values are [Br2]=2.50×10−4M , [Br−]=11.65M, [SO42−]=9.50M, and PSO2=3.50×10−5atm ? Express your answer numerically to two decimal places.