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6.1. Equilibrium - 1 (chemical Equilibrium) — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistry6.1. Equilibrium - 1 (chemical Equilibrium)1 Mark
MCQ
Write the equilibrium constant $K$ for $C{H_3}COOH + {H_2}O = {H_3}{O^ + } + C{H_3}CO{O^ - }$
  • A
    $K = \frac{{[{H_3}{O^ + }][{H_2}O]}}{{[C{H_3}CO{O^ - }][C{H_3}COOH]}}$
  • ✓
    $K = \frac{{[{H_3}{O^ + }][C{H_3}CO{O^ - }]}}{{[{H_2}O][C{H_3}COOH]}}$
  • C
    $K = \frac{{[{H_3}{O^ + }][{H_2}O]}}{{[C{H_3}COOH][C{H_3}CO{O^ - }]}}$
  • D
    $K = \frac{{[{H_2}O][C{H_3}CO{O^ - }]}}{{[{H_2}O][C{H_3}COOH]}}$
✓

Answer

Correct option: B.
$K = \frac{{[{H_3}{O^ + }][C{H_3}CO{O^ - }]}}{{[{H_2}O][C{H_3}COOH]}}$
b
Equilibrium constants for this reaction is $K =\frac{\left[ H _3 O ^{+}\right]\left[ CH _3 COO ^{-}\right]}{\left[ H _2 O \right]\left[ CH _3 COOH \right]}$

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