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:
\(\begin{aligned} & \mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{PCl}_3\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]} \\ & =\frac{\left(1.2 \times 10^{-3}\right) \times\left(1.2 \times 10^{-3}\right)}{\left(0.8 \times 10^{-3}\right)} \\ & =1.8 \times 10^{-3}~ \text {mol L}^{-1} \end{aligned}\)

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