1.

Given that bond energies of H - H and Cl - Cl are 430 kJ mol^(-1) and 240 kJ mol^(-1) respectively and Delta_(r)H for HCl is -90 kJ mol^(-1). Bond enthalpy of HCl is

Answer»

290 KJ `mol^(-1)`
380 kJ `mol^(-1)`
425 kJ `mol^(-1)`
245 kJ `mol^(-1)`

SOLUTION :`(1)/(2)H_(2)-(1)/(2)Cl_(2)rarrHCl`
`DeltaH_(f)=[(1)/(2)DeltaH_(H-H)+(1)/(2)DeltaH_(CL-Cl)]-[DeltaH_(H-Cl)]`
`-90=[(1)/(2)xx430+(1)/(2)xx240]-DeltaH_(H-Cl)`
`-90=215+120-DeltaH_(H-Cl)`
`DeltaH_(H-Cl)=335+90=425 kJ//mol`


Discussion

No Comment Found

Related InterviewSolutions