is the standard cell potential for the spontaneous between these two half cells?
ID: 996396 • Letter: I
Question
is the standard cell potential for the spontaneous between these two half cells? + 3e^- rightarrow Al E = -1.66 V + e^- rightarrow Ag E = 0.80 V 2.46 V 0.86 V 0.74 V 0.00 V series of figure shows the reaction 2A = B proceedings to dynamic equilibrium: Which figure also represents equilibrium for this reaction at the same temperature?(All boxes represents the same volume.) For an exothermic reaction performed at the two different temperatures, which condition gives the higher concentration of products at equilibrium? (Assume that temperature is the only variable.) the reaction at the higher temperature the reaction at the lower temperature the concentration will be the same because K is a constant not enough information to answer Methanol (CH_3OH) can be prepared by the reaction below. The equilibrium concentrations at 250degree C were found to be [CO] = 0.0960 M, [H_2] = 0.191 M and [CH_3OH] = 0.150 M. What is K_c for the reaction? CO(g) + 2H_2(g) rightleftharpoons CH_3OH(g) 42.8 8.18 0.122 0.0233 How does this change in stoichiometric coefficients affect the equilibrium constant expression? CH_4 + 2H_2O(g) rightleftharpoons CO_2 + 4H_2(g) K_Phi = X 2CH_4(g) + 4H_2O(g) rightleftharpoons 2CO_2 + 8H_2(g) K_Phi = ? the value of K_p is X X^2 2X 4X what is the equilibrium constant expression for this reaction? C_6H_12O_6(s) + 6CO_2(g) + 6H_2O(I) K_c = [CO_2]^6/[O_2]^6 K_c = [C_6H_12O_6] [CO_2]^6/[H_2O]^6[CO_2]^6 K_c = [CO_2]^6/[C_6H_12O_6] [O_2]^6 K_c = [CO_2]^6 [H_2O]^6/[O_2]^6 [C_6H_12O_6]^6 A flask initially contains 0.100 M nitrogen and 0.240 M oxygen. The equilibrium constant, K, for the reaction is 4.08 times 10^-4 at this temperature. What is the equilibrium concentration of NO? N_2 + O_2(g) rightleftharpoons 2NO(g) 1.6 times 10^-3 M 3.1 times 10^-3 M 0.34 M 0.68 M Which chemical equations has K = K ? 2CO(g) + O_2(g) rightleftharpoons 2CO_2(g) 4Fe(s) + 3O_2(g) rightleftharpoons 2Fe_2(s) PF_3(g) + F_2(g) rightleftharpoons PF_5(g) H_2(g) + I_2(g) rightleftharpoons 2HI(g) Which equilibrium shifts to the left side(reactants) upon a decrease in volume? 3O_2(g) rightleftharpoons 2O_2(g) Fe(s) + CO_2 rightleftharpoons FeO(s) + CO(g) 2KClO_3(s) rightleftharpoons 2KCl(s) + 3O_2(g) 4NH_3(g) + 5O_2 rightleftharpoons 4NO(g) + 6H_2O(I)Explanation / Answer
23) = 1.66 + 0.80 = 2.46
24) D because A = B1/2
25) B because the reacction is exotermic less T favored the reacction
26) CO H -> COH =) K = COH /(CO xH ) = 0.150/(0.0960x0.191) = K = 8.18
27) B if the equation is multiplice x 2
28) A because the solid a and water dont afect in the reaction of equilibium only the sustance in aquos solution and water is the solvent
29) NO2 = (K N x O)1/2 = (4.08x10-4 x 0.1 x 0.240)1/2 = 3.13x10-3
30) B because its the only whit solid present
31) C because a drecreses in the V increses the P but the solid are not affect for the pressure
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