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.

Explanation

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`