Ncert Exercises & Solutions

PCl₅, PCl₃, and Cl₂ are at equilibrium at 500 K in a closed container and their concentrations are 0.8×10^-3 mol L^-1 , 1.2×10^-3 mol L^-1and 1.2×10^-3 mol L-1, respectively.
The value of K_c for the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) will be:

1.	1.8 × 10³ mol L-1	2.	1.8 × 10³
3.	1.8 × 10^-3mol L^-1	4.	0 . 55 × 10⁴

Hint: The formula of

K_{c} = \frac{[{PCl}_{3}] [{Cl}_{2}]}{[{PCl}_{5}]}

Step 1:

For the reaction,

{PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g)

The given values are as follows:

At 500 K in a closed container,

[{PCl}_{5}] = 0.8 \times 10^{- 3}

mol

L^{- 1}

[{PCl}_{3}] = 1.2 \times 10^{- 3}

mol

L^{- 1}

[{Cl}_{2}] = 1.2 \times 10^{- 3}

mol

L^{- 1}

Step 2:

K_{c} = \frac{[{PCl}_{3}] [{Cl}_{2}]}{[{PCl}_{5}]}

= \frac{(1.2 \times 10^{- 3}) \times (1.2 \times 10^{- 3})}{(0.8 \times 10^{- 3})}

= 1.8 \times 10^{- 3}

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