Question

Calculate the enthalpy change when infinitely dilute solution of `CaCl_(2)` and `Na_(2)CO_(3)` are mixed. `Delta H_(f)^(0)` for `Ca^(+2)(aq)`.
`CO_(3)^(-2)(aq)` and `CaCO_(3)` are `-129.80,-161.7-288.50 kcal mol^(-1)` respectively.

Answer 1

Correct Answer - 3
On mixing `CaCl^(2)(aq)` and `Na_(2)CO_(3)`
`CaCl_(2) + Na^(2)CO^(3) rarr CaCO_(3) + 2NaCl`
Solutions are very dilute and thus . `100%` dissociation occurs.
`Ca^(+2)(aq) + 2Cl^(-)(aq) + 2Na^(+)(aq)+CO_(3)^(-2)(aq)`
`" "rarr CaCO_(3) downarrow +Na^(+2)(aq) +2Cl^(-)(aq)`
or `" " Ca^(+2)(aq)+ CO_(3)^(-2)(aq) rarr CaCo_(3)(s)`
`therefore " "DeltaH = sumH_("products")^(0) - sumH_("reactants")^(0)`
or `" " DeltaH =Delta_(f CaCo_(3))^(0) - [DeltaH_(fCa^(+2)) DeltaH_(fCO_(3)^(-2))^(0)]`
`therefore " " Delta H^(0)` of a command `=DeltaH_("formation")^(0)`
`" " = -288.5 -(-129.80 - 161.7) = 3Kcal`