The gas A2 in the left flask allowed to react with gas B2 present in right flask as, A2(g)+B2(g)leftharpoons2AB(g) ; Kc=4at27° C . What is the concentration of AB when equilibrium is established?

Question

<img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-vzrxahmwjiepde7c.jpg" /> The gas A2 in the left flask allowed to react with gas B2 present in right flask as, A2(g)+B2(g)leftharpoons2AB(g) ; Kc=4at27° C . What is the concentration of AB when equilibrium is established? (A) 1.33M (B) 2.66M (C) 0.66M (D) 0.33M

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-vzrxahmwjiep59cj.jpg" /> Kc=([AB]2/[A2] [B2]) ( textK) textc=(4 ( textx)2/(2 - textx) (4 - textx)) â textx = (3 2/2 4) = text1.33 mole textAB ( textg) = (2 × text1 text.33/4) = text0 text.66 M

Q.
The gas $A_{2}$ in the left flask allowed to react with gas $B_{2}$ present in right flask as, $A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$ ; $Kc = 4 a t 2 7^{\circ} C$ . What is the concentration of $A B$ when equilibrium is established?

$1.33 M$

$2.66 M$

$0.66 M$

$0.33 M$

$K_{c} = \frac{[ A B ] ^{2}}{[ A _{2} ] [ B _{2} ]}$
$(K)_{c} = \frac{4 ( x ) ^{2}}{( 2 - x ) ( 4 - x )}$
$\Rightarrow x = \frac{32}{24} = 1.33 mole$
$AB (g) = \frac{2 \times 1 .33}{4} = 0 .66 M$

Q. The gas A2​ in the left flask allowed to react with gas B2​ present in right flask as, A2​(g)+B2​(g)⇌2AB(g) ; Kc=4at27∘C . What is the concentration of AB when equilibrium is established?

1.33M

2.66M

0.66M

0.33M