InterviewSolution
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InterviewSolution
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The enthlpy of formation of hypothetical `CaCl_((s))` theoretically found to be `-188 kJ mol^(-1` and the `Delta_(f)H^(Θ)` for `CaCl_(2(s))` is `-795 kJ mol^(-1)`. Calculate the `Delta_(f)H^(Θ)` for the disproportionation reaction. `2 CaCl_((s))rarrCaCl_(2(s))+Ca_((s))` |
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Answer» Given, `Delta_(f)H^(Θ) "for" CaCl=-188 kJ mol^(-1)` `Delta_(f)H^(Θ) "for" CaCl=-795 kJ mol^(-1)` `Ca_((s))+(1)/(2)Cl_(2(s)) rarr CaCI_((s)) Delta_(f)H_(1)^(Θ)=-188 kJ mol^(-1)`…(i) `Ca_((s))+Cl_(2(g))rarrCaCl_(2(s))Delta_(f)H_(2)^(Θ)=-795 kJ mol^(-1)`…(ii) To calculate `Delta_(f)H_(3)^(Θ)` for disproportionation reaction: `2CaCl_((s))rarrCa_((s))+CaCl_(2(s)), Delta_(f)H_(3)^(Θ)=?`...(iii) `Delta_(f)H_(3)^(Θ)=Delta_(f)H_(2)^(Θ)-2Delta_(f)H_(1)^(Θ)` `=-795-(2xx-188)=-419 kJ mol^(-1)` Hence, `Delta_(f)H^(Θ)` for disproportionation of `CaCl_((s))` is `-419 kJ mol^(-1)`. |
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