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Standard XII

Chemistry

Le Chatelier's Principle for Del N Lesser Than Zero

For the react...

Question

For the reaction : $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g), K_{p} = 1$ atm at $927^{o} C$ . If 20g of $C a C O_{3}$ were kept in a 10 litre vessel at $927^{o} C$ , then calculate percentage of $C a C O_{3}$ remaining at equilibrium:

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Solution

$C a C O_{3} ⇌ C a O (s) + C O_{2} (g)$
$1$ $1$ $P$
At equilibrium
$P_{C O_{2}} = K_{p}$
$P_{C O_{2}} = 1 a t m$
volume = $10$ litre Temperature $= 927^{o} C + 273$
$= 1200 k$
$p v = n R T$
$n = \frac{p v}{R T} = \frac{1 a t m \times 10 l i t r e}{0.0821 \times 1200}$
$n = 0.1015 m o l e$
Amount of $C a C O_{3}$ consumed = $0.1015 \times (M o l a r m a s s o f C a C O_{3})$
$= 0.1015 \times 100$
$= 10.15 g m$
Amount of $C a C O_{3}$ remaining $= 20 g - 10.15 g$
$= 9.85 g \approx 10 g m$

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For the reaction : CaCO3(s)⇌CaO(s)+CO2(g),Kp=1 atm at 927oC. If 20g of CaCO3 were kept in a 10 litre vessel at 927oC, then calculate percentage of CaCO3 remaining at equilibrium:

CaCO3⇌CaO(s)+CO2(g) 1 1 PAt equilibriumPCO2=KpPCO2=1atmvolume = 10 litre Temperature =927oC+273 =1200kpv=nRTn=pvRT=1atm×10litre0.0821×1200n=0.1015mole Amount of CaCO3 consumed = 0.1015×(MolarmassofCaCO3) =0.1015×100 =10.15gmAmount of CaCO3 remaining =20g−10.15g =9.85g≈10gm

For the reaction : $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g), K_{p} = 1$ atm at $927^{o} C$ . If 20g of $C a C O_{3}$ were kept in a 10 litre vessel at $927^{o} C$ , then calculate percentage of $C a C O_{3}$ remaining at equilibrium: