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Standard XII

Chemistry

First Law of Thermodynamics

At 540 K, 0...

Question

At $540 K, 0.10 m o l e$ of $P C l_{5}$ are heated in $0.8 L$ flask. The pressure of the 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

$P C l_{5}$ $⇌$ $P C l_{3}$ $+ C l_{2}$
Initial $0.1 m o l$
At eqm. $0.1 - x$ $x$ $x$
Total no. of moles at eqm. $= 0.1 + x$
$P V = n R T$
i.e. $n=dfracPVRT=1×80.0821×540=0.18$
$∴ 0.1 + x = 0.18$ or $x = 0.08$
$∴ K_{c} = \frac{(0.08 / 8) (0.08 / 8)}{(0.02 / 8)} = 0.04$
$K_{p} = K_{c} (R T)^{Δ n} = 0.04 (0.0821 \times 540)^{1} = 1.77$

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	$P C l_{5}$	$⇌$	$P C l_{3}$	$+ C l_{2}$
Initial	$0.1 m o l$
At eqm.	$0.1 - x$		$x$	$x$

At 540 K,0.10 mole of PCl5 are heated in 0.8 L flask. The pressure of the equilibrium mixture is found to be 1.0 atm. Calculate Kp and Kc for the reaction.

PCl5⇌PCl3+Cl2Initial0.1 molAt eqm.0.1−xxxTotal no. of moles at eqm. =0.1+xPV=nRTi.e. n=dfracPVRT=1×80.0821×540=0.18∴0.1+x=0.18 or x=0.08∴Kc=(0.08/8)(0.08/8)(0.02/8)=0.04Kp=Kc(RT)Δn=0.04(0.0821×540)1=1.77

At $540 K, 0.10 m o l e$ of $P C l_{5}$ are heated in $0.8 L$ flask. The pressure of the equilibrium mixture is found to be $1.0 a t m$ . Calculate $K_{p}$ and $K_{c}$ for the reaction.