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# What is melting point of benzene if $\Delta {{H}_{fusion}}=9.95KJmo{{l}^{-1}}$ and $\Delta {{S}_{fusion}}=35.7J{{K}^{-1}}mo{{l}^{-1}}$?a.) 270 Kb.) 278 Kc.) 300 Kd.) 298 K

Last updated date: 13th Jun 2024
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Hint: Here system is changing its physical state from solid to liquid by absorbing energy and changes its physical state. Now moving into a more randomized state is going to increase the entropy of the system and we have used the relation between entropy change, enthalpy of fusion and melting point to solve this question.

Complete step by step solution:
- Melting point is the temperature at which solid substance gets converted into its liquid form.
- Enthalpy of fusion is the energy that a substance requires to change its physical state from solid to liquid at melting point.
- Entropy is the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work. Because work is obtained from ordered molecular motion, the amount of entropy is also a measure of the molecular disorder, or randomness, of a system.
- When a system changes its physical state from one phase to another, entropy also changes because in gas it has maximum randomness, after that in liquid and minimum randomness in solid. In fact solids are known for their structural order.
- That is the reason why entropy order is:
Gas > liquid > solid

- Relation between entropy change and enthalpy of fusion:
$\Delta {{S}_{fusion}}=\dfrac{\Delta {{H}_{fusion}}}{{{T}_{M.P.}}}$
so entropy change is actually the ration enthalpy fusion and melting point.
${{T}_{M.P.}}=\dfrac{\Delta {{H}_{fusion}}}{\Delta {{S}_{fusion}}}$ …………………………….. Equation (1)
$\Delta {{S}_{fusion}}=35.7J{{K}^{-1}}mo{{l}^{-1}}$
$\Delta {{H}_{fusion}}=9.95KJmo{{l}^{-1}}$
$\Delta {{H}_{fusion}}=9950J{{K}^{-1}}$
Now put the value of enthalpy change and entropy change in Equation (1) with correct units
${{T}_{M.P.}}=\dfrac{9950}{35.7}K$
${{T}_{M.P.}}=278.7K$
So the melting point of benzene is approximately equal to 278 K
So, the correct answer is “Option B”.