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Chemical Equilibrium question

2021 · 24 Feb · Shift 1 · Q15
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  5. /2021 · 24 Feb · Shift 1 · Q15

Chemical Equilibrium question

2021 · 24 Feb · Shift 1 · Q15

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
The stepwise formation of [Cu(NH3)4]2+{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}[Cu(NH3​)4​]2+ is given below: JEE Main 2021 (Online) 24th February Morning Shift Chemistry - Chemical Equilibrium Question 64 English The value of stability constants K1, K2, K3 and K4 are 104, 1.58 x 103, 5 x 102 and 102 respectively. The overall equilibrium constants for dissociation of [Cu(NH3)4]2+{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}[Cu(NH3​)4​]2+ is x ×\times× 10-12. The value of x is ‾\underline{\hspace{2cm}}​. (Rounded off to the nearest integer)
Numerical answer
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Correct answer: 1

  1. Given stepwise formation constants for Cu2++NH3⇌[Cu(NH3)]2+K1=104Cu^{2+} + NH_3 \rightleftharpoons [Cu(NH_3)]^{2+} \qquad K_1=10^4Cu2++NH3​⇌[Cu(NH3​)]2+K1​=104 [Cu(NH3)]2++NH3⇌[Cu(NH3)2]2+K2=1.58×103[Cu(NH_3)]^{2+} + NH_3 \rightleftharpoons [Cu(NH_3)_2]^{2+} \qquad K_2=1.58\times 10^3[Cu(NH3​)]2++NH3​⇌[Cu(NH3​)2​]2+K2​=1.58×103 [Cu(NH3)2]2++NH3⇌[Cu(NH3)3]2+K3=5×102[Cu(NH_3)_2]^{2+} + NH_3 \rightleftharpoons [Cu(NH_3)_3]^{2+} \qquad K_3=5\times 10^2[Cu(NH3​)2​]2++NH3​⇌[Cu(NH3​)3​]2+K3​=5×102 [Cu(NH3)3]2++NH3⇌[Cu(NH3)4]2+K4=102[Cu(NH_3)_3]^{2+} + NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+} \qquad K_4=10^2[Cu(NH3​)3​]2++NH3​⇌[Cu(NH3​)4​]2+K4​=102

  2. The overall formation constant for Cu2++4NH3⇌[Cu(NH3)4]2+Cu^{2+}+4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}Cu2++4NH3​⇌[Cu(NH3​)4​]2+ is β4=K1K2K3K4\beta_4=K_1K_2K_3K_4β4​=K1​K2​K3​K4​

  3. Substitute the values: β4=(104)(1.58×103)(5×102)(102)\beta_4=(10^4)(1.58\times 10^3)(5\times 10^2)(10^2)β4​=(104)(1.58×103)(5×102)(102)

  4. Multiply the numerical parts: 1.58×5=7.91.58\times 5=7.91.58×5=7.9

    Multiply powers of 10: 104×103×102×102=101110^4\times 10^3\times 10^2\times 10^2=10^{11}104×103×102×102=1011

    Hence, β4=7.9×1011\beta_4=7.9\times 10^{11}β4​=7.9×1011

  5. The question asks for the overall dissociation constant of [Cu(NH3)4]2+⇌Cu2++4NH3[Cu(NH_3)_4]^{2+} \rightleftharpoons Cu^{2+}+4NH_3[Cu(NH3​)4​]2+⇌Cu2++4NH3​ which is the reciprocal of the overall formation constant: Kdiss=1β4=17.9×1011K_{diss}=\frac{1}{\beta_4}=\frac{1}{7.9\times 10^{11}}Kdiss​=β4​1​=7.9×10111​

  6. Calculate: Kdiss=17.9×10−11K_{diss}=\frac{1}{7.9}\times 10^{-11}Kdiss​=7.91​×10−11 Kdiss≈0.1266×10−11=1.266×10−12K_{diss}\approx 0.1266\times 10^{-11}=1.266\times 10^{-12}Kdiss​≈0.1266×10−11=1.266×10−12

  7. Compare with the form x×10−12x\times 10^{-12}x×10−12 so, x≈1.266x\approx 1.266x≈1.266

  8. Rounded to the nearest integer: x=1x=1x=1

Final Answer: 1\boxed{1}1​

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