Question

# How much $$PCl_5$$ must be added to a one-litre vessel at $$250^oC$$ in order to obtain a concentration of 0.1 moles of $$Cl_2$$? $$K_c$$ for $$PCl_5\Longleftrightarrow PCl_3 + Cl_2$$ is 0.0414 mol/litre.

A
3.415 mole
B
34.15 mole
C
0.03415 mole
D
0.3415 mole

Solution

## The correct option is D 0.3415 moleThe equilibrium reaction is $$PCl_{5}(g) \Leftrightarrow PCl_{3}(g) + Cl_{2}(g)$$.Let a be the initial number of moles of $$PCl_5$$.Following table lists quantities of various species.                                                       $$PCl_5 \ \ \ \longrightarrow$$     $$PCl_3$$    +   $$Cl_2$$Initial number of moles                   $$a$$                          0              0Moles at equilibrium                     $$a-0.1$$                $$0.1$$             $$0.1$$$$\because$$ The equilibrium number of moles of chlorine is 0.1.Hence, the equilibrium concentration of $$PCl_5, PCl_3$$  and  $$Cl_2$$ are $$a-0.1, 0.1$$ and $$0.1$$ respectively.The expression for the equilibrium constant is $$K_c= \dfrac {[PCl_3][Cl_2]} {[PCl_5]}= \dfrac {0.1 \times 0.1} {a-0.1}=0.0414$$.Hence, $$a=0.3415$$ moles.Hence, option D is correct.Chemistry

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