Using the equilibrium constant Ka =
2.30×10-11 determine ∆rG° (in kJ
mol-1) for the following acidic
reaction at 298 K,
HypoIodous: HIO(aq) + H2O(l) <----> H3O+(aq) + IO-(aq)
Using the equilibrium constant Ka = 2.30×10-11 determine ∆rG° (in kJ mol-1) for the following acidic...
Hypoiodous acid (HIO) is a weak acid that dissociates in water as follows: HIO(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + IO−(aq). A 0.15 M solution of hypoiodous acid has a pH of 5.66. Calculate the acid-dissociation constant (Ka) for this acid.
The weak acid HIO has a Ka of 2.0×10−11. If a 1.7 M solution of the acid is prepared, what is the pH of the solution? The equilibrium expression is: HIO(aq)+H2O(l)⇋H3O+(aq)+IO−(aq) Report your answer with two significant figures.
Using provided data, determine AG* (in kJ) for the following reaction. 2036) ++3026) AH,(kJ/mol) S /mol) Ozle) 205.0 Ole) 143 238.82 Question 11 2 pts Use the provided information to determine the equilibrium constant at 298 K for the reaction given 2NO26) N204(8) 4 AH® (kJ/mol) 5° (J/molk) NO2(g) 33.2 239.9 N2048) 9.16 304.3 Equilibrium Constant - (Select) x 10 (Select)
Determine the equilibrium constant for the following reaction at 298 K: SO3(g) + H2O(g) → H2SO4(l) ΔG°rxn = -90.5 kJ mol-1 Group of answer choices 0.964 4.78 × 1011 7.31 × 1015 1.37 × 10-16 9.11 × 10-8
The change in enthalpy (DeltaHdegree_rxn) for a reaction is -22.8 kJ/mol. The equilibrium constant for the reaction is4.7 Times 10^3 at 298 K. What is the equilibrium constant for the reaction at 672 K ?
1 What is the equilibrium constant for a reaction at temperature 89.1 °C if the equilibrium constant at 22.6 °C is 49.93? For this reaction, ΔrH = -21.1 kJ mol-1 . 2 What is the ΔrG° for the following reaction (in kJ mol-1)? C6H12O6(s, glucose) + 6 O2 (g) ⇌6 CO2 (g)+ 6 H2O (l) 3 What is the ΔrG° for the following reaction (in kJ mol-1)? 2 NO2 (g) ⇌N2O4 (g) 4 What is the ΔrG for the following...
The change in enthalpy (AH) for a reaction is -35.7 kJ/mol. The equilibrium constant for the reaction is 1.4x109 at 298 K Part A What is the equilibrium constant for the reaction at 618 K? Express your answer using two significant figures. ΑΣφ ? K-
The change in enthalpy (?Horxn) for a reaction is -29.6 kJ/mol . The equilibrium constant for the reaction is 4.3×103 at 298 K. Part A What is the equilibrium constant for the reaction at 686 K ? Express your answer using two significant figures.
The value for the equilibrium constant for the following chemical reaction, the auto-ionization of water, is 1.0x10-14 at 298 K 2H2O(l) ßà OH-(aq) + H3O+(aq) K = 1.0x10-14 Using this information and your equilibrium identities, select the correct value for the equilibrium expression for the reaction shown below (at 298 K as well) : OH-(aq) + H3O+(aq) ßà 2H2O(l) K = ? Question 4 options: 1x10-14 0.5x10-14 2x10-14 -1x10-14 1x1014 1x10-15
The change in enthalpy (ΔHorxn) for a reaction is -36.6 kJ/mol . The equilibrium constant for the reaction is 2.0×103 at 298 K. What is the equilibrium constant for the reaction at 621 K ?