Write a reaction to describe the behavior of sr(c2h3o2)2 (s) in water. be sure to include all phases.

Sr(C₂H₃O₂)₂(s) + 2H₂O (l) ⇄ Sr(OH)₂(s) + 2C₂H₃OOH(aq)

Explanation:

Sr(C₂H₃O₂)₂ is a salt, formed by a metal cation (Sr⁺²) and an anion (C₂H₃O₂⁻). This ionic compound must ionize in water, making an equilibrium, which will react with the water equilibrium.

The equilibrium of the salt is:

Sr(C₂H₃O₂)₂(s) ⇄ Sr⁺²(aq) + C₂H₃O₂⁻(aq)

And the water equilibrium:

H₂O (l) ⇄ H⁺ (aq) + OH⁻(aq)

So, Sr⁺² must react with OH⁻ to form the hydroxide Sr(OH)₂, and C₂H₃O₂⁻ must react with H⁺ to form the acid C₂H₃OOH. Sr is a metal of group 2, so the base will be a little soluble in water, and the solid may precipitate. C₂H₃OOH is a weak acid, and soluble in water, so it will be in aqueous form. The reaction is:

Sr(C₂H₃O₂)₂(s) + 2H₂O (l) ⇄ Sr(OH)₂(s) + 2C₂H₃OOH(aq)

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

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enthalpy change for the reaction between the given volumes of the gases is :

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