In the preparation of quick lime from limestone- the reaction is- CaCO3s- CaOs - CO2gExperiments carried out between 850-C and 950-C led to a set of Kp values fitting in empirical equation ln Kp - 7-282 - 8500T- where T is the absolute temperature- If the reaction is carried out in quiet air- what minimum temperature would be predicted from this equation for almost complete decomposition of lime-

For complete decomposition of lime, the value of K_p must be 1 ln K_p=7.282 8500T K_p=1 7.282=8500T T=1167 K (minimum temp ...

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Standard XII

Chemistry

Empirical & Molecular Formula

In the prepar...

Question

In the preparation of quick lime from limestone, the reaction is:

$C a C O_{3 (s)} ⇌ C a O_{(s)} + C O_{2 (g)}$
Experiments carried out between $850^{\circ} C$ and $950^{\circ} C$ led to a set of $K_{p}$ values fitting in empirical equation $l n K_{p} = 7.282 - \frac{8500}{T}$ , where $T$ is the absolute temperature. If the reaction is carried out in quiet air, what minimum temperature would be predicted from this equation for almost complete decomposition of lime?

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Solution

For complete decomposition of lime, the value of $K_{p}$ must be $1$
$ln K_{p} = 7.282 - \frac{8500}{T}$
$⟹ K_{p} = 1$
$7.282 = \frac{8500}{T}$
$T = 1167$ $K$ (minimum temperature for complete reaction)
In $^{o} C = 1167 - 273 = 894^{o} C$

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For complete decomposition of lime, the value of Kp must be 1lnKp=7.282−8500T⟹Kp=17.282=8500TT=1167 K (minimum temperature for complete reaction)In oC=1167−273=894oC

For complete decomposition of lime, the value of $K_{p}$ must be $1$
$ln K_{p} = 7.282 - \frac{8500}{T}$
$⟹ K_{p} = 1$
$7.282 = \frac{8500}{T}$
$T = 1167$ $K$ (minimum temperature for complete reaction)
In $^{o} C = 1167 - 273 = 894^{o} C$