One mole of ethanol is treated with one mole of ethanoic acid at `25^(@)C`. Half of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be

One mole of ethanol is treated with one mole of ethanoic acid at 25^@C . One -fourth of the acid changes into ester at equilibrium . The equilibrium constant for the reaction will be

One mole of pure ethyl alcohol was treated with one mole of pure acetic acid at 25^(@)C One -third of the acid changes into ester at equilibrium . The equilibrium constant for the reaction will be:

One mole of ethyl alcohol was treated with one mole of acetic acid at 25^@C. 2/3 of the acid changes into ester at equilibrium Calculate the equilibrium constant for the reaction :

One mole of pure ethyl alcohil was treated with one mole of pure acetic acid at 25^(@)C One -third of the acid changes into ester at equilibrium . The equilibrium constant for the reaction will be:

One mole of ethyl alcohol (C_(2)H_(5)OH) was treated with one mole of acetic acid at 25^(0)C . (2)/(3) of the acid changes in to ester at equilibrium.The equilibrium constant for the reaction is

4 moles of A are mixed with 4 moles of B. At equilibrium for the raction A+BhArrC+D , 2 moles of C and D are formed. The equilibrium constant for the reaction will be

4 moles of A are mixed with 4 moles of B. At equilibrium for the raction A+BhArrC+D , 2 moles of C and D are formed. The equilibrium constant for the reaction will be

5 mole of X are mixed with 3 moles of Y. At equilibrium for the reaction, X + Y rarr Z 2 moles of Z are formed. The equilibrium constant for the reaction will be