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2CaSO4 (s) ˆˆ ˆˆ ˆˆˆˆ 2Ca0(s) + 2SO2(g) + O2(g), ΔH > 0
Above 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(Assume thatsolid CaSO4 is present in the container in each case).
  • a)
    moles of CaO(s) will increase with the increase in temperature
  • b)
    If tile volume of the container is doubled at equilibrium then partial pressure of SO2(g) will change at new equilibrium.
  • c)
    If the volume of the container is halved partial pressure of 02(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.
Correct answer is option 'A,C,D'. Can you explain this answer?
Verified Answer
2CaSO4 (s) 2Ca0(s) + 2SO2(g) + O2(g), H 0Above equilibrium is esta...
(A) As reaction is endothermic therefore it will go in the forward direction hence moles of Ca0 will increase.
                                (C) With the increase or decrease of volume particle pressure of the gases will remain same.
                                (D) Due to the addition of inest gas at constant pressure reaction will proceede in the direct in which move number of gaseous moles are formed.
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2CaSO4 (s) 2Ca0(s) + 2SO2(g) + O2(g), H 0Above equilibrium is esta...
Understanding the Reaction
The reaction involves the decomposition of calcium sulfate (CaSO4) into calcium oxide (CaO), sulfur dioxide (SO2), and oxygen (O2) at high temperature (1600 K). The equilibrium established can be influenced by changes in temperature, volume, and addition of gases.
Option A: Moles of CaO(s) will increase with the increase in temperature
- According to Le Chatelier’s principle, increasing the temperature favors the endothermic direction of the reaction.
- Since the decomposition of CaSO4 into CaO, SO2, and O2 is endothermic, raising the temperature increases the production of CaO.
Option B: Doubling the volume will change the partial pressure of SO2(g)
- When the volume is doubled, the partial pressure of the gases decreases due to increased volume.
- The system will shift to counteract the change, producing more SO2, thus altering its partial pressure. Hence, this option is incorrect.
Option C: Halving the volume will keep the partial pressure of O2(g) the same
- When the volume is halved, the partial pressures of the gases initially increase.
- However, the equilibrium will shift to the left (favoring reactants) to decrease the pressure. Since O2 is produced in a 1:2 ratio, its partial pressure will stabilize, making this option correct.
Option D: Adding two moles of He gas at constant pressure will increase the moles of CaO(s)
- Adding an inert gas like He at constant pressure does not change the system's equilibrium position.
- The partial pressures of the reacting gases remain unchanged, allowing the reaction to shift towards producing more CaO.
Conclusion
Thus, options A, C, and D are correct based on the principles of chemical equilibrium and gas laws.
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2CaSO4 (s) 2Ca0(s) + 2SO2(g) + O2(g), H 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(Assume thatsolid CaSO4 is present in the container in each case).a)moles of CaO(s) will increase with the increase in temperatureb)If tile volume of the container is doubled at equilibrium then partial pressure of SO2(g) will change at new equilibrium.c)If the volume of the container is halved partial pressure of 02(g) at new equilibrium will remain samed)If two moles of the He gas is added at constant pressure then the moles of CaO(s) will increase.Correct answer is option 'A,C,D'. Can you explain this answer?
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