Under similar conditions enthalpy of freezing is exactly opposite to

Under similar conditions enthalpy of freezing is exactly opposite to
  1. enthalpy of fusion
  2. enthalpy of vaporization
  3. enthalpy of solution
  4. enthalpy of atomization

Solution

For the reaction, $\mathrm{H}_2 \mathrm{O}_{(\mathrm{s})} [\text { freezing }]{\text { fusion }} \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}$ Under similar conditions of $0^{\circ} \mathrm{C}$ and 1 atm pressure, $\Delta_{\mathrm{fus}} \mathrm{H}$ is $+6.01 \mathrm{~kJ} \mathrm{~mol}^{-1}$, whereas $\Delta_{\text {free }} \mathrm{H}$ is $-6.01 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Thus, enthalpy of freezing is exactly opposite to enthalpy of fusion.

Asked in: MHT CET 2024 (15 May Shift 1)

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