The dissociation of CaCO3 takes place as per the equation- CaCO3s - CaOs - CO2g- - H - 110 kJ- If the reaction is carried out in a closed vessel- the pressure of CO2 is increased and reaches a constant value when-

The correct option is A Temperature is increasedThe equilibrium reaction is CaCO_3(s) ⇔ CaO(s) + CO_2(g); Δ H = 110 kJ .Positive value of the enthalpy c ...

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Byju's Answer

Standard XII

Chemistry

Le Chatelier's Principle for Del N Lesser Than Zero

The dissociat...

Question

The dissociation of $C a C O_{3}$ takes place as per the equation,

$C a C O_{3} (s) \Leftrightarrow C a O (s) + C O_{2} (g); Δ H = 110 k J$ .

If the reaction is carried out in a closed vessel, the pressure of $C O_{2}$ is increased and reaches a constant value when:

Temperature is increased

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Temperature is decreased

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Volume of vessel is increased

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Amount of

C a C O_{3}

is decreased

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Solution

The correct option is A Temperature is increased
The equilibrium reaction is $C a C O_{3} (s) \Leftrightarrow C a O (s) + C O_{2} (g); Δ H = 110 k J$ .

Positive value of the enthalpy change indicates that the reaction is endothermic.

$∴$ The forward reaction is favored by increasing temperature.

Thus, when the temperature is increased, more carbon dioxide is formed which increases the pressure of carbon dioxide until it reaches a constant value.

Hence, option $A$ is correct.

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Similar questions

Solids $C a C O_{3}$ and CaO and gaseous $C O_{2}$ are placed in a vessel and allowed to reach equilibrium

$C a O (s) + C O_{2} (g) ⇌ C a C O_{3} (s) Δ H = - 180 k J m o l^{- 1}$

The quantity of CaO in the vessel could be increased by:

Q. Assertion :The equilibrium (given below) attained in a closed vessel remains unaltered by the addition of

C a C O_{3} (s)

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

Reason: Heterogenous reaction is unaffected by solid and liquid.

20.0

grams of

C a C O_{3} (s)

were placed in a closed vessel, heated & maintained at

727^{o}

C under equilibrium

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

and it is found that

75 %

C a C O_{3}

was decomposed. What is the value of

K_{p}

? The volume of the container was

15

litre.

Q. In the equilibrium,

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

, the number of moles of

C O_{2}

is increased by

Q. Given the enthalpy of formation of

C O_{2}

(g) is

- 94.0

KJ, of CaO(s) is

- 152

KJ, and the enthalpy of the reaction

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

is 42 KJ, the enthalpy of formation of

C a C O_{3} (s)

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The dissociation of CaCO3 takes place as per the equation,CaCO3(s)⇔CaO(s)+CO2(g);ΔH=110kJ. If the reaction is carried out in a closed vessel, the pressure of CO2 is increased and reaches a constant value when:

The dissociation of $C a C O_{3}$ takes place as per the equation,

$C a C O_{3} (s) \Leftrightarrow C a O (s) + C O_{2} (g); Δ H = 110 k J$ .

If the reaction is carried out in a closed vessel, the pressure of $C O_{2}$ is increased and reaches a constant value when: