In the preparation of CaO from CaCO 3 using the equilibrium CaCO 3 s - CaO s - CO 2 g K p is expressed as log K p -7-282-8500- T forthe complete decomposition of CaCO 3- the temperature in Celsius to be used isA- 1167B- 894C- 8500D- 850

The correct option is B 894 For complete decomposition, K_p=1 ∴ log 1=7.282 = 8500/T i.e., 7.282 = 8500/T, (∵ log 1=0) or, T = 8500/7.282 = 1167 K =1167 273 ...

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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

In the prepar...

Question

In the preparation of CaO from $C a C O_{3}$ using the equilibrium $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g) K_{p}$ is expressed as log $K_{p} = 7.282 - \frac{8500}{T}$ for the complete decomposition of $C a C O_{3}$ , the temperature in Celsius to be used is

1167

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894

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8500

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850

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Solution

The correct option is B 894
For complete decomposition, $K_{p} = 1$
$∴ l o g 1 = 7.282 = \frac{8500}{T}$
i.e., $7.282 = \frac{8500}{T}, (∵ l o g 1 = 0)$
or, $T = \frac{8500}{7.282} = 1167 K$
$= 1167 - 273^{\circ} C = 894^{\circ} C$

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In the preparation of CaO from CaCO3 using the equilibrium CaCO3(s)⇌CaO(s)+CO2(g)Kp is expressed as log Kp=7.282−8500T for the complete decomposition of CaCO3, the temperature in Celsius to be used is

The correct option is B 894 For complete decomposition, Kp=1 ∴log 1=7.282=8500T i.e., 7.282=8500T,(∵log 1=0) or, T=85007.282=1167 K =1167−273∘C=894∘C

In the preparation of CaO from $C a C O_{3}$ using the equilibrium $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g) K_{p}$ is expressed as log $K_{p} = 7.282 - \frac{8500}{T}$ for the complete decomposition of $C a C O_{3}$ , the temperature in Celsius to be used is

The correct option is B 894
For complete decomposition, $K_{p} = 1$
$∴ l o g 1 = 7.282 = \frac{8500}{T}$
i.e., $7.282 = \frac{8500}{T}, (∵ l o g 1 = 0)$
or, $T = \frac{8500}{7.282} = 1167 K$
$= 1167 - 273^{\circ} C = 894^{\circ} C$