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When 1g liquid naphthalene (C(10)H(8)) s...

When `1g` liquid naphthalene `(C_(10)H_(8))` solidifies, `149J` of heat is evolved. Calculate the enthalpy of fusion of naphthalene. (Write answer in Joules)

Text Solution

Verified by Experts

The correct Answer is:
19072
Molecular mass of naphthalene, `C_(10)H_(8) = 10 xx 12 + 8 xx 1 = 128`
For the solidification reaction
`C_(10)H_(8)(l) to C_(10)H(s)`

Heat evolved when `1` g of naphthalene solidifies = `149 J`
Heat evolved when `128 g` of naphthalene solidify = `149 xx 128 = 19072 J`
i.e.,
`C_(10)H_(8)(l) to C_(10)H_(8)(s) Delta_(fus) H^(@) = -19072 J`

For the fusion reaction,
`C_(10)H_(8)(s) to C_(10)H_(8)(l)`

This reaction is the reverse of the above solidfication reaction so that
`Delta_(fus)H^(@) = -Delta_(solid) H^(@)`
Thus `Delta_(fus)H^(@) = 19072 J or = 19.072 kJ`
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