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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
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B
34.15 mole
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C
0.03415 mole
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D
0.3415 mole
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Solution

The correct option is D 0.3415 mole
The 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              0

Moles 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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