What is ∆G° for the reaction I₂(g) → 2 I(g) at 25.0 °C if K = 6.1 x 10⁻²²? (R = 8.314 J/mol･K)

The first reaction in Glycolysis is the phosphorylation of Glucose:Pi + Glucose → Glucose 6-Phosphate + waterThis is a thermodynamically unfavorable process with a ∆G0’ = +13.8 kJ/mol. In a liver cell at37 0C the concentrations of both phosphate and glucose are normally maintained at about 0.005 M each. What would be the equilibrium concentration of Glucose 6-Phosphate according to the above data? Note: In the biochemical standard state ​[H2O] = 1.0

Determine ∆G° for a reaction when ∆G = -129.5 kJ/mol and Q = 0.043 at 298 K. (R = 8.314 J/mol ･ K)

Determine ∆G° for a reaction when ∆G = -117.2 kJ/mol and Q = 23 at 298 K. (R = 8.314 J/mol ･ K)

Determine ∆G° for a reaction when ∆G = -188.0 kJ/mol and Q = 0.035 at 223 K. (R = 8.314 J/mol ･ K)

This reaction must have a ∆G<0 at SATP:   Xe(g)+F2(g)→XeF2(g)   2Na(s)+Cl2(g)→2NaCl(s)   2NaF(s)+I2(s)→2NaI(s)+F2(g)   MgSO4(s)→Mg(s)+S(s)+2O2(g)

For butane, the ∆H° of vaporization is 22.40 kJ/mol and the ∆S° of vaporization is 82.3 J/mol･K. At 1.00 atm and 203.0 K, what is the ∆G° of vaporization for butane, in kJ/mol?

For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -4.5 kJ/mol?

For butane, the ∆H° of vaporization is 22.4 kJ/mol and the ∆S° of vaporization is 82.3 J/mol･K. At 1.00 atm and 201.0 K, what is the ∆G° of vaporization for butane, in kJ/mol?