Question

The enthalpy of formation of hypothetical `CaCl(s)` is theoretically found to be `-188 k J mol^(-1)` and `Delta_(f) H^(@)` for `CaCl_(2) (s) is -795 k J mol^(-1)` . Calculate `Delta_(f) H^(@)` (in kJ `mol^(-1)`) for the disproportionation reaction .
`2 CaCl(s) to CaCl_(2) + Ca (s)`

Answer 1

Correct Answer - 419
`Delta_(f) H^(@)` for the given reaction
`= Delta_(f) H^(@) ("products") - Delta_(f) H^(@) ("reactants")`
`= Delta_(f) H^(@) (CaCl_(2)) + Delta_(f) H^(@) - 2 Delta_(f) H^(@) (CaCl)`
`= -795 + 0 - 2 xx - 188 k J mol^(-1)`
`= -419 k J mol^(-1)`