1. Tardigrade
  2. Question
  3. Chemistry
  4. AB dissociates as 2 AB ( g ) leftharpoons 2 A ( g )+ B 2( g ) When the initial pressure of AB is 500 mm Hg, the pressure becomes 625 mm Hg when the equilibrium is attained. Calculate KP for the reaction assuming volume remains constant.

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Q. $AB$ dissociates as
$2 AB ( g ) \rightleftharpoons 2 A ( g )+ B _{2}( g )$
When the initial pressure of $AB$ is $500\, mm\, Hg$, the pressure becomes $625\, mm\, Hg$ when the equilibrium is attained. Calculate $K_{P}$ for the reaction assuming volume remains constant.

Equilibrium

Solution:

$\underset{(500-p) mm }{2 AB} \rightleftharpoons \underset{P}{2 A} +\underset{P/2}{B _{2}}$

Equilibrium pressures

$(500-p)+p+p / 2=625$ mm

$p / 2=125$ mm

$p=250$ mm

$p_{A B}=(500-250)=250$ mm

$p_{A}=p=250$ mm

$p_{B_{2}}=p / 2=\frac{250}{2}$ mm

$=125$ mm

$K_{p}=\frac{p_{ B _{2}} \times p_{ A }^{2}}{p_{ AB }^{2}}$

$=\frac{125 \times(250)^{2}}{(250)^{2}}$

$=125$ mm