If two gases AB2 and B2C are mixed, following equilibrium are readily established.AB2(g)+B2C(g)⇌AB3(g)+BC(g)BC(g)+B2C(g)⇌B3C2(g)If the reaction is started only with AB2 and B2C , then which of the following is nearly true at equilibrium, considering initial moles to be same?

AB2aa−x;+B2Caa−x−y;⇌AB3-x;+BC-x−y;BCx−y;+B2Ca−x−y;⇌B3C2y⇒x>y or AB3eq>B3C2eq

If two gases AB2 and B2C are mixed, following equilibrium are readily established.AB2(g)+B2C(g)⇌AB3(g)+BC(g)BC(g)+B2C(g)⇌B3C2(g)If the reaction is started only with AB2 and B2C , then which of the following is nearly true at equilibrium, considering initial moles to be same?

AB2aa−x;+B2Caa−x−y;⇌AB3-x;+BC-x−y;BCx−y;+B2Ca−x−y;⇌B3C2y⇒x>y or AB3eq>B3C2eq

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If two gases ${AB}_{2}$ and $B_{2} C$ are mixed, following equilibrium are readily established.
$\begin{array}{l} A B_{2} (g) + B_{2} C (g) ⇌ A B_{3} (g) + B C (g) \\ B C (g) + B_{2} C (g) ⇌ B_{3} C_{2} (g) \end{array}$
If the reaction is started only with ${AB}_{2}$ and $B_{2} C$ , then which of the following is nearly true at equilibrium, considering initial moles to be same?

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${{\{A B_{3}]}_{e q} <∣ B C]}_{e q}$

${[A B_{2}]}_{e q} = {[B_{2} C]}_{e q}$

${[A B_{3}]}_{e q} > {[B_{3} C_{2}]}_{e q}$

${{[A B_{3}]}_{e q} =∣ B C]}_{e q}$

answer is C.

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Detailed Solution

$\underset{\underset{a - x;}{a}}{A B_{2}} + \underset{\underset{a - x - y;}{a}}{B_{2} C} ⇌ \underset{\underset{x;}{-}}{A B_{3}} + \underset{\underset{x - y;}{-}}{B C}$

$\underset{x - y;}{B C} + \underset{a - x - y;}{B_{2} C} ⇌ \underset{y}{B_{3} C_{2}}$

$\Rightarrow x > y$ or ${[A B_{3}]}_{e q} > {[B_{3} C_{2}]}_{e q}$

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If two gases AB2 and B2C are mixed, following equilibrium are readily established.AB2(g)+B2C(g)⇌AB3(g)+BC(g)BC(g)+B2C(g)⇌B3C2(g)If the reaction is started only with AB2 and B2C , then which of the following is nearly true at equilibrium, considering initial moles to be same?

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