In the preparation of CaO from Ca CO 3 using the equilibrium Ca CO 3 s- CaOs- CO 2 g Kp is expressed as- log K p -7-282- 8500 Tfor complete decomposition of Ca CO 3 the temperature in celsius to be used is-

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Standard XII

Chemistry

Reaction Quotient

In the prepar...

Question

In the preparation of $C a O$ 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 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 $894$
In the preparation of $C a O$ 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 complete decomposition of $C a {C O}_{3}$ the temperature in celsius to be used is $894$ .
For complete decomposition of $C a {C O}_{3}$
$K_{p} = 1$
$log K_{p} = l o g 1 = 0$
$log K_{p} = 7.282 - \frac{8500}{T}$
$0 = 7.282 - \frac{8500}{T}$
$7.282 = \frac{8500}{T}$
$T = 1167$ K
$T = 1167 - 273 = 894^{o} C$ .

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In the preparation of CaO from CaCO3 using the equilibrium CaCO3(s)⇌CaO(s)+CO2(g) Kp is expressed as:logKp=7.282−8500Tfor complete decomposition of CaCO3 the temperature in celsius to be used is:

In the preparation of $C a O$ 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 complete decomposition of $C a {C O}_{3}$ the temperature in celsius to be used is: