JEE MAIN - Chemistry (2019 - 12th January Morning Slot - No. 4)
In a chemical reaction,
_12th_January_Morning_Slot_en_4_1.png)
the initial concentration of B was 1.5 times of the concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant (K) for the aforesaid chemical reaction is -
the initial concentration of B was 1.5 times of the concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant (K) for the aforesaid chemical reaction is -
16
1
1/4
4
Explanation
_12th_January_Morning_Slot_en_4_3.png)
At equilibrium [A] = [B]
$$ \Rightarrow $$ a - x = 1.5a - 2x
$$ \Rightarrow $$ x = 0.5a
Kc = $${{{{\left[ C \right]}^2}\left[ D \right]} \over {\left[ A \right]{{\left[ B \right]}^2}}}$$
= $${{{{\left( a \right)}^2}\left( {0.5a} \right)} \over {\left( {0.5a} \right){{\left( {0.5a} \right)}^2}}}$$ = 4
Comments (0)
