Questions: A CaSOA system is at equilibrium at 25°C. CaSO4(s) ⇌ Ca^2+(aq) + SO4^2-(aq) Which direction will the reaction shift if the equilibrium system is warmed to 40°C ? Ksp = [Ca^2+][SO4^2-]

A CaSOA system is at equilibrium at 25°C.

CaSO4(s) ⇌ Ca^2+(aq) + SO4^2-(aq)

Which direction will the reaction shift if the equilibrium system is warmed to 40°C ?

Ksp = [Ca^2+][SO4^2-]
Transcript text: A CaSOA system is at equilibrium at $25^{\circ} \mathrm{C}$. \[ \mathrm{CaSO}_{4}(\mathrm{~s}) \rightleftharpoons \mathrm{Ca}^{2+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \] Which direction will the reaction shift if the equillbrium system is warmed to $40^{\circ} \mathrm{C}$ ? \[ \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{SO}_{4}^{2-}\right] \]
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Solution

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Solution Steps

Step 1: Analyze the given information

The dissolution of calcium sulfate (CaSO₄) is given by the equation: \[\mathrm{CaSO}_{4}(\mathrm{~s}) \rightleftharpoons \mathrm{Ca}^{2+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\] We are also told that the system is at equilibrium at 25°C and is then warmed to 40°C. We are asked to determine the direction the reaction will shift.

Step 2: Determine the nature of the reaction

The dissolution of CaSO₄ is an endothermic reaction. This means that heat is absorbed during the dissolution process, and can be considered a reactant: \[\mathrm{CaSO}_{4}(\mathrm{~s}) + \text{Heat} \rightleftharpoons \mathrm{Ca}^{2+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\]

Step 3: Apply Le Chatelier's principle

Le Chatelier's principle states that if a change in conditions is applied to a system at equilibrium, the system will shift in a direction that relieves the stress. In this case, increasing the temperature from 25°C to 40°C adds heat to the system. Since heat is a "reactant" in the endothermic dissolution of CaSO₄, the system will shift to the right to consume the added heat.

Final Answer

The reaction will shift to the right, favoring the formation of more $\text{Ca}^{2+}$ and $\text{SO}_4^{2-}$ ions. \( \boxed{\text{Right}} \)

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