Question

Consider the reaction  PCl_3(g) + Cl_2(g)  ⇋   PCl_5(g)   K_c = 0.040 at 450^oC. If a reaction is initiated with 0.40 mole of Cl₂ and 0.40 mole of PCl₃ in a 2.0 liter container, what is the equilibrium concentration of PCl5 in the same system?

Answer 1

$K c ​ =20= \frac{(0.4+x)}{(0.2−x)(0.3−x)}$

$20= \frac{0.4+x}{0.06−0.5x+x ^2}$

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∴20x² −11x+0.8=0

∴x=0.08