1.

The standard enthalpies of formation of HCl(g), H(g) and Cl(g) are -92.2, 217.7 and 121.4 kJ mol^(-1) respectively. The bond dissociation enthalpy of HCl is :

Answer»

`+431.3 kJ`
236.9 kJ
`-431.3 kJ`
339.1 kJ

Solution :`HCL(g) LT H(g) + CL(g)`
`DELTA H = [Delta_(f)H(H(g)) + Delta_(f)H Cl(g)] - Delta_(f)H (HCl(g))`
`= 217.7 + 121.4 - (- 92.2) = 431.3 kJ`


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