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Le Chatelier's Principle for Del N Equals to 0

2CaSO4 s ⇌ 2C...

Question

$2 C a S O_{4} (s) ⇌ 2 C a O (s) + 2 S O_{2} (g) + O_{2} (g), Δ H > 0$
Above equilibrium is established by taking sufficient amount of CaSO $_{4}$ (s) in a closed container at 1600 K. Then which of the following may be correct option(s)?(Assume that solid CaSO $_{4}$ is represent in the container in each case)

Moles of CaO(s) will increase with the increase in temperature.

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If the volume of the container is doubled at equilibrium then partial pressure of SO

_{2}

(g) will change at new equilibrium.

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If the volume of the container is halved partial pressure of O

_{2}

(g) at new equilibrium will remain same.

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If two moles of the He gas is added at constant pressure then the moles of CaO(s) will increase.

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Solution

The correct options are
A Moles of CaO(s) will increase with the increase in temperature.
C If the volume of the container is halved partial pressure of O $_{2}$ (g) at new equilibrium will remain same.
D If two moles of the He gas is added at constant pressure then the moles of CaO(s) will increase.
The given reaction is :-
$2 C a {S O}_{4} (s) ⇌ 2 C a O (s) + 2 {S O}_{2} (g) + O_{2} (g); △ H > 0$
The given reaction is endothermic reaction reaction as $△ H > 0$ , so after increasing the temperature, by Le Chatelier's principle, reaction will proceed in forward direction.
So, on increasing temperature moles of $C a O$ will increase.
Now we have
$K p = p S O_{2}$
$K_{o}$ is dependent only on temperature.
So $K_{p}$ will remain constant even if volume of the container is halved .As a result partial pressure of $S O_{2}$ at equilibrium remains same.
When the inert gas is added to the system at equilibrium at constant pressure then the volume increases so the reaction will shift itself toward backwards side in the direction of volume decreasing. It can be concluded as a result of Le chatelier's principle.

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Similar questions

$2 C a S O_{4} (s) ⇌ 2 C a O (s) + 2 S O_{2} (g) + O 2 (g), Δ H > 0$

Above equilibrium is established by taking sufficient amount of $C a S O_{4} (s)$ in a closed container at 1600 K. Then, which of the following may be correct options?
(Assume that solid $C a S O_{4}$ is present in the container in each case).

2 C a S O_{4} (s) ⇌ 2 C a O (s) + 2 S O_{2} (g) + O_{2} (g), Δ H > 0

Above equilibrium is established by taking a sufficient amount of

C a S O_{4}

(s) in a closed container at 1600 K. Then which of the following may be the correct option(s)?

(Assume that solid

C a S O_{4}

is present in the container in each case)

2 C a S O_{4} (s) ⇌ 2 C a O (s) + 2 S O_{2} (g) + O_{2} (g), Δ H > 0

If the above shown equilibrium is established by taking sufficient amount of

C a S O_{4} (s)

in a closed container at 1600K, then which of the following is(are) correct? (Assume that solid

C a S O_{4}

is present in the container in each case)

Consider the following statements:

(I) An increase in pressure (caused by decrease in volume) at equilibrium results in increase in molar concentration of each gaseous substance involved.

(II) An increase in pressure (caused by decrease in volume) at equilibrium results in increase in no. of moles of each gaseous substance involved.

(III) For the reaction

A (s) ⇌ B (g) + C (g)

at constant temperature, total pressure at equilibrium (of

B (g)

C (g)

) is not affected by changing the volume of container.

(IV)

H_{2} O (l) ⇌ H_{2} O (g)

on temperature increase,

K_{p}

of this reaction is increased.

Tick the correct statement(s)

Q. If

K_{P}

of an equilibrium

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

is 1.5 atm at

20^{o} C

then the moles of CaO formed if excess

C a C O_{3} (s)

is being placed in a 1 lt of closed container containing

C O_{2}

gas at 0.5 atm pressure at same temperature is:

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2CaSO4(s)⇌2CaO(s)+2SO2(g)+O2(g),ΔH>0Above equilibrium is established by taking sufficient amount of CaSO4(s) in a closed container at 1600 K. Then which of the following may be correct option(s)?(Assume that solid CaSO4 is represent in the container in each case)