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JEE Main 202124 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2021 · 24 February · Morning Shift · Chemistry · Chemical Equilibrium

The stepwise formation of [Cu ( NH _ 3 ) _ 4 ] ^ 2 + is given...

The stepwise formation of [Cu(NH3)4]2+{\left[Cu{\left({NH}_{3}\right)}_{4}\right]}^{2 +} is given below:

Cu2++NH3⇌K1[Cu(NH3)]2+{Cu}^{2 +} + {NH}_{3}\overset{{K}_{1}}{⇌}{\left[Cu\left({NH}_{3}\right)\right]}^{2 +}

[Cu(NH3)]2++NH3⇌K2[Cu(NH3)2]2+{\left[Cu\left({NH}_{3}\right)\right]}^{2 +} + {NH}_{3}\overset{{K}_{2}}{⇌}{\left[Cu{\left({NH}_{3}\right)}_{2}\right]}^{2 +}

[Cu(NH3)2]2++NH3⇌K3[Cu(NH3)3]2+{\left[Cu{\left({NH}_{3}\right)}_{2}\right]}^{2 +} + {NH}_{3}\overset{{K}_{3}}{⇌}{\left[Cu{\left({NH}_{3}\right)}_{3}\right]}^{2 +}

[Cu(NH3)3]2++NH3⇌K4[Cu(NH3)4]2+{\left[Cu{\left({NH}_{3}\right)}_{3}\right]}^{2 +} + {NH}_{3}\overset{{K}_{4}}{⇌}{\left[Cu{\left({NH}_{3}\right)}_{4}\right]}^{2 +}

The value of stability constants K1,K2,K3{K}_{1}, {K}_{2}, {K}_{3} and K4{K}_{4} are 104,1.58×103,5×102{10}^{4}, 1 . 58 \times {10}^{3}, 5 \times {10}^{2} and 102{10}^{2} respectively. The overall equilibrium constants for dissociation of [Cu(NH3)4]2+{\left[Cu{\left({NH}_{3}\right)}_{4}\right]}^{2 +} is x×10−12.x \times {10}^{- 12} . The value of xx is _____ (Rounded off to the nearest integer)

Correct answer

0

Step-by-step solution

Cu2++NH3⇌K1[Cu(NH3)]2+{Cu}^{2 +} + {NH}_{3}\overset{{K}_{1}}{⇌}{\left[Cu\left({NH}_{3}\right)\right]}^{2 +}

[Cu(NH3)]2++NH3⇌K2[Cu(NH3)2]2+{\left[Cu\left({NH}_{3}\right)\right]}^{2 +} + {NH}_{3}\overset{{K}_{2}}{⇌}{\left[Cu{\left({NH}_{3}\right)}_{2}\right]}^{2 +}

[Cu(NH3)2]2++NH3⇌K3[Cu(NH3)3]2+{\left[Cu{\left({NH}_{3}\right)}_{2}\right]}^{2 +} + {NH}_{3}\overset{{K}_{3}}{⇌}{\left[Cu{\left({NH}_{3}\right)}_{3}\right]}^{2 +}

[Cu(NH3)3]2++NH3⇌K4[Cu(NH3)4]2+{\left[Cu{\left({NH}_{3}\right)}_{3}\right]}^{2 +} + {NH}_{3}\overset{{K}_{4}}{⇌}{\left[Cu{\left({NH}_{3}\right)}_{4}\right]}^{2 +}

Cu2++4NH3⇌K[Cu(NH3)4]2+{Cu}^{2 +} + 4{NH}_{3}\overset{K}{⇌}{\left[Cu{\left({NH}_{3}\right)}_{4}\right]}^{2 +}

So,

K=K1×K2×K3×K4K = {K}_{1} \times {K}_{2} \times {K}_{3} \times {K}_{4}

=104×1.58×103×5×102×102= {10}^{4} \times 1 .58 \times {10}^{3} \times 5 \times {10}^{2} \times {10}^{2}

K=7.9×1011K = 7 .9 \times {10}^{11}

Where K→K \to Equilibrium constant for formation of [Cu( NH 3)4]{\left[Cu{\left({\text{ NH }}_{3}\right)}_{4}\right]}^{2 +}<br/>Soequilibriumconstant(K′)<br />So equilibrium constant \left({K}^{'}\right) for dissociation
of[Cu(NH3)4]2+is1K\text{of}{\left[Cu{\left({NH}_{3}\right)}_{4}\right]}^{2 +}\text{is}\dfrac{1}{K}

K′=1K{K}^{'} = \dfrac{1}{K}

K′=17.9×1011{K}^{'} = \dfrac{1}{7 . 9 \times {10}^{11}}

=1.26×10−12=(x×10−12)= 1 . 26 \times {10}^{- 12} = \left(x \times {10}^{- 12}\right)

So the value of x=1.26x = 1 . 26

OMROMR Ans =1= 1 (After rounded off to the nearest integer)

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