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Chemical Equilibrium question

2009 · Q118
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Chemical Equilibrium question

2009 · Q118

NEETChemistryChemical EquilibriumMCQ+4 / −1
The dissociation constants for acetic acid and HCN at 25oC are 1.5 ×\times× 10−-−5 and 4.5 ×\times× 10−-−10 respectively. The equilibrium constant for the equilibrium
CN−-− + CH3COOH ⇌\rightleftharpoons⇌ HCN + CH3COO−-− would be
  1. A
    3.0 ×\times× 10−-−5
  2. B
    3.0 ×\times× 10−-−4
  3. C
    3.0 ×\times× 104
  4. D
    3.0 ×\times× 105
View written solutionFree

Correct answer: C

CH3COOH ⇌\rightleftharpoons⇌ CH3COO–

  • H+,   K1 = 1.5 ×\times× 10−-−5

    ⇒\Rightarrow⇒ K1 = [CH3COO−][H+][CH3COOH]{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {C{H_3}COOH} \right]}}[CH3​COOH][CH3​COO−][H+]​ = 1.5 ×\times× 10−-−5

    HCN ⇌\rightleftharpoons⇌ CN–
  • H+,     K2 = 4.5 × 10–10

    K2 = [CN−][H+][HCN]{{\left[ {C{N^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {HCN} \right]}}[HCN][CN−][H+]​ = 4.5 × 10–10

    CN−-− + CH3COOH ⇌\rightleftharpoons⇌ HCN + CH3COO−-−

    K = [HCN][CH3COO−][CN−][CH3COOH]{{\left[ {HCN} \right]\left[ {C{H_3}CO{O^ - }} \right]} \over {\left[ {C{N^ - }} \right]\left[ {C{H_3}COOH} \right]}}[CN−][CH3​COOH][HCN][CH3​COO−]​

    ⇒\Rightarrow⇒ K = K1K2{{{K_1}} \over {{K_2}}}K2​K1​​ = 1.5×10−54.5×10−10{{1.5 \times {{10}^{ - 5}}} \over {4.5 \times {{10}^{ - 10}}}}4.5×10−101.5×10−5​

    = 3.33 × 104

    ≃\simeq≃ 3.0 × 104
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