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Chemical Equilibrium question

2006 · Shift 0 · Q5

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Chemical Equilibrium
2006 · Shift 0 · Q5

Chemical Equilibrium question

2006 · Shift 0 · Q5

JEE MainChemistryChemical EquilibriumMCQ+4 / −1

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel

PCl_5

(g)

\leftrightharpoons

PCl_3

(g) +

Cl_2

(g) If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of

PCl_5

is x, the partial pressure of

PCl_3

will be

A
$\left( {{x \over {x + 1}}} \right)P$
B
$\left( {{2x \over {1 - x}}} \right)P$
C
$\left( {{x \over {x - 1}}} \right)P$
D
$\left( {{x \over {1 - x}}} \right)P$

View written solutionFree

Correct answer: A

Assume initial moles

Let the initial moles of $PCl_5$ be $1$ .

Reaction: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$

If the degree of dissociation is $x$ , then at equilibrium:

Moles of $PCl_5 = 1-x$
Moles of $PCl_3 = x$
Moles of $Cl_2 = x$

Total moles at equilibrium

$n_{\text{total}}=(1-x)+x+x=1+x$

Mole fraction of $PCl_3$

$y_{PCl_3}=\frac{x}{1+x}$

Partial pressure of $PCl_3$

Partial pressure $=$ mole fraction $\times$ total pressure:

$p_{PCl_3}=\frac{x}{1+x}P$

Match with options

$p_{PCl_3}=\left(\frac{x}{x+1}\right)P$

This matches Option A.

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