JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
The enthalpy change on freezing of \(1\, mol\) of water at \(5\,^oC\) to ice at \(-5\,^oC\) is .....\( kJ\, mol^{-1}\) (Given \({\Delta _{fus}}H = 6\, kJ\, mol^{-1}\) at \(0\,^oC\), \(C_p(H_2O, l) =75.3\, J\, mol^{-1} \, K^{-1}\) , \(C_p(H_2O, s) = 36.8\, J\, mol^{-1} \, K^{ -1}\) )
- A \(5.44\)
- B \(5.81\)
- C \(6.56\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(6.56\)
Step-by-step Solution
Detailed explanation
In order to calculate the enthalpy change for \(H_2O\) at \(5\,^oC\) to ice at \(-\,5\,^oC\) , we need to calculated the enthalpy change of all the transformation involved in the process. \((a)\) Energy change of \(1\,mol\), \(H_2O\,(l)\), at \(5\,^oC\) \(\to \,1\,mol\),…
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