Calculate AG° for the reaction A + B +>C+D at 25°C when the equilibrium conditions are [A] = 12 mM, [B] = 12 mM, [C] = 9.8 mM, [D] = 9.8 mM. A) -1 J/mol B) -100 J/mol C) -1 kJ/mol D) -10 kJ/mol E) none of the above

Consider the following equilibrium process at 686. °℃: CO₂(g) + H₂(g) CO(g) + H₂O(g) The equilibrium concentrations of the reacting species are [CO]=0.0560 M, [H₂] = 0.0400 M, [CO₂]=0.0830 M, and [H₂O]=0.0430 M. Part 1 of 5 Calculate K for the reaction at 686. °C. Round your answer to 3 significant digits. с Kc- Part 2 of 5 x10 If we add CO₂ to increase its concentration to 0.510 M, what will the concentrations of all the gases be when equilibrium is reestablished? Round your answer to 3 significant digits. [CO₂] = M x10 X Ś

Keq for the reaction below is 0.680. Initially the concentration of carbon monoxide is 0.500 mol/L and chlorine is 1.00 mol/L. Determine the equilibrium concentrations of all entities. (This reaction will actually proceed to the LEFT!)  COCl2 <---> CO(g) + Cl2(g)

The equilibrium constant for a reaction is 0.38 at 25 °C. What is the value of AG° (kJ/mol) at this temperature? More information is needed. O 200 O 4.2 2.4 -4.2 « Previous Ne MacBook Air F4 FS 5 8 9 H K C >

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Need help understanding questions 6

Calculate AG°for each of the following reactions from the equilibrium constant at the temperature given. Reaction Temperature (°C) 2000 N₂ (g) + O₂(g) →→→ 2NO (g) H₂ (g) + 12 (g) → 2HI (g) 400 CO₂ (g) + H₂ (9) → CO (g) + H₂O (g) 980 CaCO3 (s) → CaO (s) + CO2 (g) 900 Ko value 4.1 x 10-4 50.0 1.67 1.04

You are given the following reaction at equilibrium in water: NH4CI(s) NH4*(aq) + CI(aq) AH= +33.4 k]/mol a) What change do you observe upon the addition of NaOH? b) What change do you observe upon the addition of heat? c) How is this observation consistent with Le Chatlier's Principle? Explain.

Consider the following system at equilibrium where AH° = 16.1 kJ, and K = 0.00650, at 298 K: 2NOBr(g) → 2NO(g) + Br₂ (g) When 0.36 moles of NO(g) are added to the equilibrium system at constant temperature: The value of K O increases decreases remains the same The value of Qc O is greater than K O is equal to Ke O is less than Ke The reaction must Orun in the forward direction to reestablish equilibrium run in the reverse direction to reestablish equilibrium O remain in the current position, since it is already at equilibrium The concentration of NO will increase decrease O remain the same