1.0 mol of $P C l_{3} (g)$ and 2.0 mol of $C l_{2} (g)$ were placed in a 3 L flask and heated to 400 K. When equilibrium was established, only 0.70 mol of $P C l_{3} (g)$ remained. What is the value of equilibrium constant for the reaction:
$P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$ at 400 K?

0.25

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1.31

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0.76

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2.6

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Solution

The correct option is C 0.76
The following table lists the number of moles of different species.

$P C l_{3}$ $C l_{2}$ $P C l_{5}$

Initial 1 2 0

change -x -x x

Equilibrium 1 - x 2 - x x

But at equilibrium, $[P C l_{3}] = 0.7 = 1 - x \Rightarrow x = 0.3$
Thus, the number of moles of $P C l_{3}, C l_{2}$ and $P C l_{5}$ are 0.7, 1.7 and 0.3 respectively.
The molar concentrations of $P C l_{3}, C l_{2}$ and $P C l_{5}$ are $\frac{0.7}{3} M, \frac{1.7}{3} M$ and $\frac{0.3}{3} M$ respectively.
So, According to definition of equilibrium constant.
So, $K_{c} = \frac{(0.3) (3)}{(1.7) (0.7)} = 0.76$

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Similar questions

Q. 1 mole of

P C l_{3} (g)

with 2 moles of

C l_{2} (g)

were placed in a 3 lit vessel and heated to 400K. When equilibrium was established, only

0.7

moles of

P C l_{3}

remained. What is the value of equilibrium constant

K_{c}

for the reaction

P C l_{3} + C l_{2} \Leftrightarrow P C l_{5}

The equilibrium constant for the reaction given below is $2.0 \times 10^{- 7} a t 300 K$ .
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