The equilibrium composition for the reaction is - 0-20PCl3 - 0-10Cl2g0-40 moles- litre- PCl5gIf 0-20 mole of Cl2 is added at same temp- Find equilibrium concentration of PCl5 KC - 20-

The correct option is D 0.48 #160; Pcl_3 + Cl_2_(g) ↔ Pcl_5_(g) #10; 0.2 x #160; #160; #160; #160; 0.3 x #160; #1 ...

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Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

The equilibri...

Question

The equilibrium composition for the reaction is -
$P C l_{3} 0.20 + C l_{2} (g) 0.10 ⇌ P C l_{5} (g) 0.40 m o l e s / l i t r e$
If $0.20 m o l e$ of $C l_{2}$ is added at same temp. Find equilibrium concentration of $P C l_{5} (K_{C} = 20)$ .

0.48

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0.38

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0.56

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1.20

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Solution

The correct option is D $0.48$
$P c l_{3} + {C l_{2}}_{(g)} \leftrightarrow {P c l_{5}}_{(g)}$

$0.2 - x$ $0.3 - x$ $0.4 + x$

$K_{c} = 20 = \frac{(0.4 + x)}{(0.2 - x) (0.3 - x)}$

$∴ 20 = \frac{0.4 + x}{0.06 - 0.5 x + x^{2}}$

$∴ 20 x^{2} - 11 x + 0.8 = 0$

$∴ x = 0.08$

$∴$ Concentration of $P c l_{5} = 0.4 + 0.08 = 0.48$

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Similar questions

The equilibrium composition for the reaction is -

$P C l_{3} + C l_{2} ⇌ P C l_{5}$

0.20 0.10 0.40 mole/litre

If 0.20 mole of $C l_{2}$ is added at same temp. Find equilibrium concentration of $P C l_{5}$ $K_{C} = 20$

The equilibrium composition in 1 litre container for the reaction is
$P C l_{3} + C l_{2} ⇌ P C l_{5} 0.20 0.08 0.40$

0.22 mole of $C l_{2}$ is added at the same temperature and a new equilibrium is established. If the sum of the new equilibrium moles of $P C l_{3}, C l_{2} a n d P C l_{5}$ is p, then value of p is