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in Chemistry by (92.8k points)
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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
A. 290 kJ `mol^(-1)`
B. 380 kJ `mol^(-1)`
C. 425 kJ `mol^(-1)`
D. 245 kJ `mol^(-1)`

1 Answer

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Best answer
Correct Answer - C
`(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`

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