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Standard XII

Chemistry

Reaction Quotient

At 540 K, ...

Question

At $540 K$ , $0.10 m o l e$ of $P {C l}_{5}$ are heated in an $8$ litre flask. The pressure of equilibrium mixture is found to be $1.0 a t m$ . Calculate $K_{p}$ and $K_{c}$ for the reaction.

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Solution

$P {C l}_{5} ⇌ P {C l}_{3} (g) + {C l}_{2} (g)$
At equi,(mole) $(0.1 - x$ $x$ $x$
total number of moles, $n = (0.1 - x) + x + x = (0.1 + x)$
$P V = n R T$
$1 \times 8 = (0.1 + x) \times 0.082 \times 540$
or $x = 0.08$
$K_{c} = \frac{[P {C l}_{3}] [{C l}_{2}]}{[P {C l}_{5}]} = \frac{x^{2}}{(0.1 - x) \times 8} = 4 \times 10^{- 2} m o l L^{- 1}$
$K_{p} = K_{c} {(R T)}^{Δ n} = K_{c} (R T) [Δ n = + 1] = 4 \times 10^{- 2} \times 0.082 \times 540 = 1.77 a t m$

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At 540 K, 0.10 mole of PCl5 are heated in an 8 litre flask. The pressure of equilibrium mixture is found to be 1.0 atm. Calculate Kp and Kc for the reaction.

PCl5⇌PCl3(g)+Cl2(g)At equi,(mole) (0.1−x x xtotal number of moles, n=(0.1−x)+x+x=(0.1+x)PV=nRT1×8=(0.1+x)×0.082×540or x=0.08Kc=[PCl3][Cl2][PCl5]=x2(0.1−x)×8=4×10−2molL−1Kp=Kc(RT)Δn=Kc(RT)[Δn=+1]=4×10−2×0.082×540=1.77atm

At $540 K$ , $0.10 m o l e$ of $P {C l}_{5}$ are heated in an $8$ litre flask. The pressure of equilibrium mixture is found to be $1.0 a t m$ . Calculate $K_{p}$ and $K_{c}$ for the reaction.