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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Chemistry
Chapter
Chemical Equilibrium

JIPMER 2015 · Chemistry · Chemical Equilibrium

For the following equilibrium (omitting charges) I. M+Cl M Cl, K_...

For the following equilibrium (omitting charges) I. M+Cl⟶MCl,Keq=β1M+\mathrm{Cl} \longrightarrow M \mathrm{Cl}, K_{\mathrm{eq}}=\beta_1 II. MCl+Cl⟶MCl2,Keq =β2M C l+\mathrm{Cl} \longrightarrow M C l_2, K_{\text{eq }}=\beta_2 III. MCl2+Cl⟶MCl3,Keq =β3M \mathrm{Cl}_2+\mathrm{Cl} \longrightarrow \mathrm{MCl}_3, K_{\text{eq }}=\beta_3 IV. M+3Cl⟶MCl3,Keq=KM+3 \mathrm{Cl} \longrightarrow M \mathrm{Cl}_3, K_{\mathrm{eq}}=K then relationship between K,β1,β2K, \beta_1, \beta_2 and β3\beta_3 is

Options

  1. AK=β1β2β3K=\beta_1 \beta_2 \beta_3
  2. Blog⁡K=log⁡β1+log⁡β2+log⁡β3\log K=\log \beta_1+\log \beta_2+\log \beta_3
  3. CpK=pβ1+pβ2+pβ3\mathrm{p}_{\mathrm{K}}=\mathrm{p}_{\beta_1}+\mathrm{p}_{\beta_2}+\mathrm{p}_{\beta_3}
  4. DAll of the above

Correct answer

D. All of the above

Step-by-step solution

On adding I, II and III, we get IV ∴K=β1β2β3\therefore \quad K=\beta_1 \beta_2 \beta_3 or, log⁡K=log⁡β1+log⁡β2+log⁡β3\log K=\log \beta_1+\log \beta_2+\log \beta_3 Or −log⁡K=−log⁡β1−log⁡β2−log⁡β3PK=Pβ1+Pβ2+Pβ3-\log K=-\log \beta_1-\log \beta_2-\log \beta_3 P_{\mathrm{K}}=P_{\beta_1}+P_{\beta_2}+P_{\beta_3} (where,px=−log⁡x)\left.p_x=-\log x\right)
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