1. Where is the position of equilibrium for reactions with thefollowing equilibr

Question

1. Where is the position of equilibrium for reactions with thefollowing equilibrium constants?    a. 500 at 80oC    b. 1.2 x 10-2 @20oC    c. 6.5 x 103 @25oC    d. 0.247 at 100oC 2. The equilibrium constant for the following reaction is0.212@100oC when[NO2] = 0.20M.     What is the equilibrium[N2O4]? 1. Where is the position of equilibrium for reactions with thefollowing equilibrium constants?    a. 500 at 80oC    b. 1.2 x 10-2 @20oC    c. 6.5 x 103 @25oC    d. 0.247 at 100oC 2. The equilibrium constant for the following reaction is0.212@100oC when[NO2] = 0.20M.     What is the equilibrium[N2O4]?

Explanation / Answer

2NO2 <....................>N2O4 so K =[N2O4]eq/[NO2]2eq when K = 500 then [N2O4]eq =500[NO2]2eq so the poistion of equlibrium istowards right. when K = 0.012 then [N2O4]eq =0.012[NO2]2eq so the poistion of equlibriumis towards left. when K = 6500 then [N2O4]eq =6500[NO2]2eq so the poistion of equlibrium istowards right when K= 0.247 then [N2O4]eq =0.247[NO2]2eq so the poistion of equlibriumis towards left. --------------- 2)   K =[N2O4]eq/[NO2]2eq        0.212 =[N2O4]eq/ (0.20M)2           [N2O4]eq =0.212*(0.20M)2   = 0.00848M           [N2O4]eq =0.212*(0.20M)2   = 0.00848M

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