4. A reaction has an equ reaction at this temperature. (25 points) has an equilibrium constant...
The equilibrium constant for a reaction is 0.48 at 25 c. What is the value of AG (k/mol) at this temperature Seleccione una: a. 4.2 b. 1.8 d. 150 e. More information is needed. The equilibrium constant for a reaction is 0.48 at 25 cl What is the value of AG /mob at this temperature Seleccione una: a. 4.2 b. 18 c. 4.2 d. 150 e. More information is needed.
The equilibrium constant for a reaction is 0.35 at 25 °C. What is the value of AG* (kj/mol) at this temperature? A. 2.6 B.-4.2 C. 220 D.4.2 E. More information is needed.
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...
4.(4 points) What is the value of equilibrium constant for the system shown below at 25°C in terms of Ka HNO2 and Kb C5H5N? No2 (ag) + C6H5NH*(ag) = C6H5N(ag) + HNO2(ag) (а) Ка+ Kb (b) Ka - Kb (c) Ka/ Kb (d) Kw / (Ka Kb) (e) -Ka Kb) ОООО
A reaction is shown to have an equilibrium constant K of 45. was run at 100 °C? What is the AG for this reaction if it
Using the reduction potentials given, calculate the equilibrium constant, K, at 20 degrees C for the reaction Using the reduction potentials given, calculate the equilibrium constant, K, at 25°C for the reaction, 33 3+ Ag (aa) t Fe(a)Ag) Fe (aq) +0.77 V +0.80 V A Ag+(aq) + e- ← a. 1.66 b. 6.4 c. 3.2 d. 6.1 x 10-4 e. 1.6 x 104 Rank the following compounds according to increasing solubility in water. K” is a less than sign) 34...
An equilibrium reaction of the formula HA ↔ H+ + A- has an equilibrium constant of 13.5 and a [H+] = 0.500 M. What is the value of [HA]? (Show work) A. 13.5 M B. 54.0 M C. 0.250 M D. 0.018 M
A reaction has an equilibrium constant of 8x103 at 298 K. At 716 K, the equilibrium constant is 0.77. Find AHºrxn for the reaction. Enter your answer numerically in kJ to 4 decimal places.
Deriving concentrations from data The equilibrium constant, K, of a reaction at a particular temperature is detemined by the concentrations or pressures of the reactants and products at equilibrium. In Part A, you were given the equilibrium pressures, which could be plugged directly into the formula for K. In Part B however, you will be given initial concentrations and only one equilibrium concentration. You must use this data to find all three equilibrium concentrations before you can apply the formula...
Calculate the equilibrium constant for the following reaction at 25 °C and 150°C. 11. (2) NH CI() NH()HC1(g) NH,C1 (s) -202.87 NH( -16.45 HC1(8) -95.30 AG (kJ-mo1-1) AG [-16.45(-95.0 -202.87] 91.12 kJ At 298 K: K#e*(- Gran/RT) e*(91120/(8.314 x 298) 1.1 At 423 K: K= e"(-aGrin/RT)-e*(91 120/(8.314 x 423) = 5.6 × 10-16 10-12 Calculate the equilibrium constant for the following reaction at 25 °C and 150°C. 11. (2) NH CI() NH()HC1(g) NH,C1 (s) -202.87 NH( -16.45 HC1(8) -95.30 AG...