Question bank/JEE Main/Chemistry

Phosphorus pentachloride dissociates as follows in a closed
reaction vessel.
PCl_5(g) [IMAGE] 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 PCl3
will be:
gksrh gS
PCl_5(g) [IMAGE] PCl_3(g) + Cl_2(g)
fn vfHkfØ;k feJ.k ij dqy nkc P gS rFkk PCl_5 sfV x

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

Show answer and solution▾

Given: PCI5 (g)[IMAGE] PCI3(g) + CI2(g)
t = 0 1 0 0
teq 1-x x x
Total number of moles = 1 - x + x + x = 1 + x
Thus partial pressure of PCI3 = $\left( \frac{x}{1 + x} \right)$ P.
fn;k gqvk gS PCI5 (g)[IMAGE] PCI3(g) + CI2(g)
t = 0 1 0 0
teq 1-x x x
dqy = 1 - x + x + x = 1 + x
vr PCI3 = $\left( \frac{x}{1 + x} \right)$ P.

Source: Grass-G / CH-26 Chemistry Paper 1 15 Jan.docx