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What is ΔrH° for the following phase change? LiCl(s) → LiCl(l)

LiCl(s)
-408.27 ΔH°f (kJ/mol-rxn)

LiCl(l)
-390.76 ΔH°f (kJ/mol-rxn)

Respuesta :

sebassandin

Answer:

[tex]\Delta _rH=17.51\frac{kJ}{mol}[/tex]

Explanation:

Hello!

In this case, since the enthalpy change for any process is computed by subtracting the enthalpy of the final state and the enthalpy of the initial state, for the given phase change, we subtract the enthalpy of the liquid (final state) and the enthalpy of the solid (initial state) considering this a melting process:

[tex]\Delta _rH=-390.76\frac{kJ}{mol} -(-408.27\frac{kJ}{mol} )\\\\\Delta _rH=17.51\frac{kJ}{mol}[/tex]

Which makes sense because this process absorbs energy.

Best regards!

pstnonsonjoku

The enthalpy change for the sublimation of LiCl(s) → LiCl(l) is 17.51 kJ/mol.

The enthalpy change for the phase change is given by;

ΔH°fproducts - ΔH°freactants.

From the available information in the question;

ΔH°fLiCl(s)   = -408.27 kJ/mol

ΔH°fLiCl(l) = -390.76 kJ/mol

Hence;

ΔH°reaction = (-390.76 kJ/mol) - (-408.27 kJ/mol)

ΔH°reaction = 17.51 kJ/mol

Learn more: https://brainly.com/question/11897796

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