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Chemical Equilibrium question

2021 · 26 Aug · Shift 2 · Q21
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  5. /2021 · 26 Aug · Shift 2 · Q21

Chemical Equilibrium question

2021 · 26 Aug · Shift 2 · Q21

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
The reaction rate for the reaction [PtCl4]2−[PtCl_4]^{2-}[PtCl4​]2−+ H2OH_2OH2​O ⇌\rightleftharpoons⇌[Pt(H2O)Cl3]−[Pt(H_2O)Cl_3]^-[Pt(H2​O)Cl3​]− + Cl−Cl^-Cl− was measured as a function of concentrations of different species. It was observed that −d[[PtCl4]2−]dt=4.8×10−5[[PtCl4]2−]−2.4×10−3[[Pt(H2O)Cl3]−][Cl−]{{ - d\left[ {{{\left[ {PtC{l_4}} \right]}^{2 - }}} \right]} \over {dt}} = 4.8 \times {10^{ - 5}}\left[ {{{\left[ {PtC{l_4}} \right]}^{2 - }}} \right] - 2.4 \times {10^{ - 3}}\left[ {{{\left[ {Pt({H_2}O)C{l_3}} \right]}^ - }} \right]\left[ {C{l^ - }} \right]dt−d[[PtCl4​]2−]​=4.8×10−5[[PtCl4​]2−]−2.4×10−3[[Pt(H2​O)Cl3​]−][Cl−]. where square brackets are used to denote molar concentrations. The equilibrium constant Kc = ‾\underline{\hspace{2cm}}​ . (Nearest integer)
Numerical answer
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Correct answer: 0

  1. Write the reaction and rate law

The reaction is:

[PtCl4]2−+H2O⇌[Pt(H2O)Cl3]−+Cl−[PtCl_4]^{2-} + H_2O \rightleftharpoons [Pt(H_2O)Cl_3]^- + Cl^-[PtCl4​]2−+H2​O⇌[Pt(H2​O)Cl3​]−+Cl−

Given rate expression:

−d[[PtCl4]2−]dt=4.8×10−5[[PtCl4]2−]−2.4×10−3[[Pt(H2O)Cl3]−][Cl−]-\frac{d[[PtCl_4]^{2-}]}{dt} = 4.8\times 10^{-5}[[PtCl_4]^{2-}] - 2.4\times 10^{-3}[[Pt(H_2O)Cl_3]^-][Cl^-]−dtd[[PtCl4​]2−]​=4.8×10−5[[PtCl4​]2−]−2.4×10−3[[Pt(H2​O)Cl3​]−][Cl−]

This has the standard reversible form:

net forward rate=kf[A]−kb[B][C]\text{net forward rate} = k_f[A] - k_b[B][C]net forward rate=kf​[A]−kb​[B][C]

so we identify:

kf=4.8×10−5k_f = 4.8\times 10^{-5}kf​=4.8×10−5 kb=2.4×10−3k_b = 2.4\times 10^{-3}kb​=2.4×10−3
  1. Use equilibrium condition

At equilibrium, net rate is zero:

0=kf[[PtCl4]2−]−kb[[Pt(H2O)Cl3]−][Cl−]0 = k_f[[PtCl_4]^{2-}] - k_b[[Pt(H_2O)Cl_3]^-][Cl^-]0=kf​[[PtCl4​]2−]−kb​[[Pt(H2​O)Cl3​]−][Cl−]

Therefore,

kf[[PtCl4]2−]=kb[[Pt(H2O)Cl3]−][Cl−]k_f[[PtCl_4]^{2-}] = k_b[[Pt(H_2O)Cl_3]^-][Cl^-]kf​[[PtCl4​]2−]=kb​[[Pt(H2​O)Cl3​]−][Cl−]

Rearranging,

[[Pt(H2O)Cl3]−][Cl−][[PtCl4]2−]=kfkb\frac{[[Pt(H_2O)Cl_3]^-][Cl^-]}{[[PtCl_4]^{2-}]} = \frac{k_f}{k_b}[[PtCl4​]2−][[Pt(H2​O)Cl3​]−][Cl−]​=kb​kf​​

Since H2OH_2OH2​O is the solvent, it is omitted from KcK_cKc​. Hence,

Kc=[[Pt(H2O)Cl3]−][Cl−][[PtCl4]2−]=4.8×10−52.4×10−3K_c = \frac{[[Pt(H_2O)Cl_3]^-][Cl^-]}{[[PtCl_4]^{2-}]} = \frac{4.8\times 10^{-5}}{2.4\times 10^{-3}}Kc​=[[PtCl4​]2−][[Pt(H2​O)Cl3​]−][Cl−]​=2.4×10−34.8×10−5​
  1. Calculate
Kc=4.82.4×10−5+3=2×10−2=0.02K_c = \frac{4.8}{2.4}\times 10^{-5+3} = 2\times 10^{-2} = 0.02Kc​=2.44.8​×10−5+3=2×10−2=0.02
  1. Nearest integer
0.02≈00.02 \approx 00.02≈0

So the nearest integer value of KcK_cKc​ is:

0\boxed{0}0​
  1. Compare with stored answer

Stored correct answer = 000.

Our derived answer matches it.

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