InterviewSolution
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InterviewSolution
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The enthalpy of formation of hypothetical CaCl(s) is found to be - 180 kJ `mol^(-1)` and that of `CaCI_(2)` (s) is -800 kJ `mol^(-1)`. Calculate `Delta_(f)H^(@)` for the disproportionation reaction: `2CaCI(s) to CaCI_(2)(s) + Ca(s)` |
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Answer» The available data is : (i) `Ca(s) + 1//2 CI_(2) (g) to CaCI(s)`, `Delta_(f) H^(@) = - 180 kJ mol^(-1)` (ii) `Ca(s) + CI_(2) (g) to CaCI_(2)(s),` `Delta_(f)H^(@) = - 800 kJ mol^(-1)` multiply eqn. (i) by 2 and subract from eqn. (ii) `Ca(s) + CI_(2) to CaCL(s),` `Delta_(f)H^(@) = - 800 kJ mol^(-2)` `2Ca (s) + CI_(2)(g) to 2CaCI(s),` `Delta_(f)H^(@) = - 360 kJ mol^(-1)` `overline("Subtract" : 2CaCI(s) to CaCI_(2) (s) + Ca (s) , Delta_(f)H^(@) = - 420 kJ mol^(-1))` |
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