If 4.6 gm of ethyl alcohol and 6.0 gm of acetic acid is kept at a constant temperature until equilibrium was established, 2.0 gm of unused acetic acid were present. What is the Kc?
Correct answer: C. 4.0
- A. 2.0
- B. 3.0
- C. 4.0
- D. 5.0
Explanation
Initially the moles of ethyl alcohol are:4.6/46=0.1 mol and acetic acid are:6/60=0.1 mol.In the end, we have 2gm of acetic acid that means 2/60=0.0333 mol of acetic acid is left or conversely speaking 0.0667 mol of acetic acid reacted to produce the same amount of ethyl acetate and water.So now Kc can be calculated as we now at the final stage of reaction we have 0.0333 mol of acetic acid and thus 0.0333 mol ethyl alcohol present as reactants, as molar ratio is same in the equation. We also know that since 0.0667 mol reacted so ethyl acetate is present in the same amount as water.Kc=[0.0667]2/[0.0333]2=4.0
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