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Given: i. H(g) +CI(g) rarr HCI(g) DeltaH =- 431 kJ ii. HCI(g) +aq rarr H^(o+)(aq) +CI^(Theta) (aq) DeltaH =- 75.1 kJ iii. H(g) rarr H^(o+) (aq)+e^(-) DeltaH = 1317 kJ iv. CI(g) +e^(-) rarr cI^(Theta)(g) DeltaH =- 354 kJ a. Calculate the enthalpy of hydration of HCI H^(o+)(g) +CI^(Theta)(g) +aq rarr H^(o+)(aq) +CI^(Theta)(aq) b. Calculate teh enthalpy of hydration of CI^(Theta) ions if enthalpy of hydration of H^(o+) is zero. |
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Answer» SOLUTION :a. `H^(o+)(g) +CI^(Theta) (g) +aq rarr H^(o+)(aq) +CI^(Theta)(aq) DELTAH = ?` `DeltaH = DeltaH_(1) +DeltaH_(2) - DeltaH_(3) - DeltaH_(4)` `=- 431 - 75.1 - 1317 - (-354)` `=- 1469.1 kJ` b. `CI^(Theta)(g)+aq rarr CI^(Theta) (aq), DeltaH_(5) = ?` `H^(o+) (g) +aq rarr H^(o+) (aq), DeltaH_(6) = 0` Given: `H^(o+)(g) +CI^(Theta)(g)+aq rarr H^(o+) (aq) +CI^(Theta) (aq)` `DeltaH =- 1469.1 kJ` `DeltaH = DeltaH_(5) +DeltaH_(6)` `:. DeltaH_(5) = DeltaH - DeltaH_(6)` `= - 1469.1 - 0 =- 1469.1 kJ` |
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