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Standard XIII

Chemistry

Derivation of Kp and Kc

When limeston...

Question

When limestone is heated, quicklime is formed according to the equation.
$C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$
The experiment was carried out in the temperature range $800 - 900^{o} C,$ Equilibrium constant $K_{p}$ follows the relation. $l o g K_{p} = 7.282 - \frac{8500}{T}$
where, T is temperature in Kelvin. At what temperature the decomposition will give $C O_{2} (g)$ at 1 atm?

{894.11}^{o} C

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{794.66}^{o} C

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{635.45}^{o} C

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{987.18}^{o} C

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Solution

The correct option is A ${894.11}^{o} C$
The given equation for decompositon of Quicklime $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$
The experiment was carried out in the temperature range $800 - 900^{o} C .$ Equilibrium constant $K_{p}$ follows the relation,
$l o g K_{p} = 7.282 - \frac{8500}{T}$
where, T is temperature in Kelvin.

The decomposition gives $C O_{2} (g)$ at 1 atm
$K_{p} = P_{C O_{2}} = 1$
$l o g K_{p} = 7.282 - \frac{8500}{T}$
$l o g 1 = 7.282 - \frac{8500}{T}$
$0 = 7.282 - \frac{8500}{T}$
$7.282 = \frac{8500}{T}$
$T = {1167.26}^{o} C$
$T = 1167.26 - {273.15}^{o} C = {894.11}^{o} C$
The temperature is ${894.11}^{o} C$

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Similar questions

Q. When limestone is heated, quicklime is formed according to the equation,

C a {C O}_{3} (s) ⇌ C a O (s) + C O_{2} (s)

The experiment was carried out in the temperature range

{800 - 900}^{o} C

. Equilibrium constant

K_{p}

follows the relation,

log K_{p} = 7.282 - 8500 / T

where,

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is temperature in Kelvin. At what temperature the decomposition will give

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Q. When limestone is heated, quicklime is formed according to the equation.

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)

The experiment was carried out in the temperature range

800 - 900^{o} C,

Equilibrium constant

K_{p}

follows the relation.

l o g K_{p} = 7.282 - \frac{8500}{T}

where, T is temperature in Kelvin. At what temperature the decomposition will give

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C a C O_{3 (s)} ⇌ C a O_{(s)} + C O_{2 (g)}

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Derivation of Kp and Kc

Standard XIII Chemistry

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