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`
Calculate the enthalpy change when infinitely dilute solutions of `CaCl_(2)` and `Na_(2)CO_(3)` are mixed. `Delta_(f)H^(@)` for `Ca_((aq.))^(2+),CO_(3(aq.))^(2-)` and `CaCO_(3(s))` are `-129.80,-161.65` and `288.46kcal//mol` respectively.
Rosey Umar Mukherjee
Asked: 2 years ago2022-11-07T10:04:25+05:30
2022-11-07T10:04:25+05:30In: General Awareness
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.
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.
`CO_(3)^(-2)(aq)` and `CaCO_(3)` are `-129.80,-161.7-288.50 kcal mol^(-1)` respectively.
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Yasmin Deshmukh
Asked: 2 years ago2022-10-29T01:57:57+05:30
2022-10-29T01:57:57+05:30In: General Awareness
Calculate the enthalpy change when infinitely dilute solutions of `CaCl_(2)` and `Na_(2)CO_(3)` are mixed. `Delta_(f)H^(@)` for `Ca_((aq.))^(2+),CO_(3(aq.))^(2-)` and `CaCO_(3(s))` are `-129.80,-161.65` and `288.46kcal//mol` respectively.
Calculate the enthalpy change when infinitely dilute solutions of `CaCl_(2)` and `Na_(2)CO_(3)` are mixed. `Delta_(f)H^(@)` for `Ca_((aq.))^(2+),CO_(3(aq.))^(2-)` and `CaCO_(3(s))` are `-129.80,-161.65` and `288.46kcal//mol` respectively.
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Correct Answer – 3
The given reaction on two solution mixing is
`CaCl_(2)+Na_(2)CO_(3) rarr CaCO_(3)+2NaCl`
Each` 100%` dissociated at infinite dilution, thus
`Ca_((aq.))^(2+)+2Cl_(aq.)^(-)+2Na_((aq.))^(+)+CO_(3(aq.))^(2-)rarr 2Na_((aq.))^(+)+2Cl_((aq.))^(-)+CaCO_(3(s))`
or `CO_(3(aq.))^(2-)+Ca_((aq.))^(2+)rarr CaCO_(3(s))`
`:. Delta_(r)H=Sigma_(Pr od uct)^(@)-Sigma_(reactant)^(@)`
`=Delta_(r)H^(@)Sigma_(CaCO_(3))^(@)-[Delta_(f)H^(@)Sigma_(Ca^(2+))+Delta_(f)H^(@)Sigma_(CO_(3)^(2-))]`
`=-288.46-(-129.80-161.65)`
`=2.99kcal~~3kcal`
Correct Answer – 4 KCAL
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) darr + 2Na^(+) (aq) + 2Cl^(-) (aq)` or
`Ca^(2+) (aq) + CO_(3)^(2-) (aq) rarr CaCO_(3) (s)`
`:. DeltaH = SigmaH_(“Products”)^(0)` or `Delta H = DeltaH_(f CaCO_(3))^(0) – [Delta H_(f ca^(2+))^(0) + Delta H_(f CO_(3)^(2))^(0)]`
`:. H^(0)` of a compound
`= H_(“Formation”)^(0) = – 288.45 – (-129.80 – 161.65) = 3 k cal`