JEE MAIN - Chemistry (2020 - 9th January Evening Slot - No. 18)
In the figure shown below reactant A
(represented by square) is in equilibrium with
product B (represented by circle). The
equilibrium constant is :
_9th_January_Evening_Slot_en_18_1.png)
1
2
8
4
Explanation
In the figure, Reactant A is represented by square.
And there are 6 squares. So at equilibrium
[A]eq = 6
In the figure, Product B is represented by circle.
And there are 11 circles. So at equilibrium
[B]eq = 11
For reaction, A ⇌ B
Keq = $${{\left[ B \right]} \over {\left[ A \right]}}$$ = $${{11} \over 6}$$ $$ \approx $$ 2
And there are 6 squares. So at equilibrium
[A]eq = 6
In the figure, Product B is represented by circle.
And there are 11 circles. So at equilibrium
[B]eq = 11
For reaction, A ⇌ B
Keq = $${{\left[ B \right]} \over {\left[ A \right]}}$$ = $${{11} \over 6}$$ $$ \approx $$ 2
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