Q8.one mole of N2 and 3 moles of PCl5 are placed in a 100l vessel heated to 227C the equilibrium pressure is 2.05 atm assuming ideal behaviour calculate the degree of dissociation for PCl5 and kp for the reaction PCl5--> PCl3 +Cl2

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${PCl}_{5} (g) \leftrightarrow {PCl}_{3} (g) + {Cl}_{2} (g) 3 mol 0 0 3 (1 - α) 3 α 3 α Total mol of gases in the vessel, n = n_{N_{2}} + n_{{PCl}_{5}} + n_{{PCl}_{3}} + n_{{Cl}_{2}} = 1 + 3 (1 - α) + 3 α + 3 α = 4 + 3 α Also, we know that PV = nRT So, from the given equilibrium pressure, n = \frac{PV}{RT} = \frac{2.05 \times 100}{0.0821 \times 500} = 5 mol Therefore, 4 + 3 α = 5 or, 3 α = 1 or, α = \frac{1}{3} = 0.333 Therefore, n_{{PCl}_{5}} = 3 (1 - 0.333) = 2 mol n_{{PCl}_{3}} = n_{{Cl}_{2}} = 3 (0.333) = 1 mol Now, p_{{PCl}_{5}} = \frac{n_{{PCl}_{5}}}{n} p_{eq} = \frac{2}{5} \times 2.05 = 0.82 atm p_{{PCl}_{3}} = p_{{Cl}_{2}} = \frac{1}{5} \times 2.05 = 0.41 atm K_{p} = \frac{(p_{{PCl}_{3}}) (p_{{Cl}_{2}})}{(p_{{PCl}_{5}})} = \frac{(0.41) (0.41)}{(0.82)} = 0.205 atm$

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Q. One mole of

N_{2}

and

3 m o l e s

P C l_{5}

are placed in a

100

litre vessel heated to

227^{\circ} C

. The equilibrium pressure is

2.05

atm. Assuming ideal behaviour, calculate degree of dissociation of

P C l_{5}

and

K_{p}

of the reaction:

P C l_{5} ⇌ P C l_{3} + C l_{2}

At $227^{o} C$ one mole of $C l_{2}$ and 3 moles of $P C l_{5}$ are placed in a 100 litre container and allowed to reach equilibrium. It is found that the equilibrium pressure is 2.05 atmospheres. Assuming ideal behaviour, the degree of dissociation 'x' for $P C l_{5}$ and $K_{p}$ for the reaction $P C l_{5}$ (g) $⇌$ $P C l_{3}$ (g) + $C l_{2}$ (g) is:

One mole of $N_{2} a n d 3$ moles of $P C l_{5}$ are placed in a 100 litre vessel heated to $227^{\circ} C$ . The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate degree of dissociation of $P C l_{5}$ for the reaction $P C l_{5} ⇌ P C l_{3} + C l_{2}$ .