Chemistry · Equilibrium involving chemical processes: Law of chemical equilibrium, equilibrium constants (Kp and Kc) and their significance, the significance of Delta G and Delta G° in chemical equilibrium
One mole of ethanol is treated with one mole of ethanoic acid at One-forth of th
One mole of ethanol is treated with one mole of ethanoic acid at \( 25 . \) One-forth of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be:
- A. \( 1 / 9 \)
- B. \( 4 / 9 \)
- C. 9
- D. \( 9 / 4 \)
Step-by-step solution
Initial moles: ethanol=1, ethanoic acid=1, ester=0, water=0. At equilibrium, one-fourth of acid (0.25 mol) reacts, producing 0.25 mol ester and 0.25 mol water. Remaining: ethanol=0.75 mol, acid=0.75 mol. Since volume constant, Kc = ([ester][water])/([ethanol][acid]) = (0.25×0.25)/(0.75×0.75) = 1/9.
Related MCQs
- for this reaction is 10. If 1,2,3,4 mole/litre of and respectively are present in a container at the direction of reaction will be:…
- The equilibrium constant for a reaction is and the reaction quotient is . For a reaction mixture, the ratio is 0.33 This means that:…
- For the following equilibrium, in gaseous phase, is of the total volume when equilibrium is set up. Hence, percent of dissociation of is:…
- For } \rightleftharpoons \boldsymbol{X} \boldsymbol{Y}_{(\boldsymbol{g})}+\boldsymbol{Y}_{(\boldsymbol{g})}, \) intially 1 mole each of and …
- 2 moles of when heated in a closed vessels of 2 litre capacity at equilibrium of dissociated in and What is the value of the equilibrium con…