Respuesta :

Answer:

Entropy change = -168.3 J/mol-K

Explanation:

The entropy change is expressed as:

[tex]DeltaS = \frac{DeltaH}{T}[/tex]

For HBr, the enthalpy of condensation = -19.27 kJ/mol

Temperature T = -67 C = 273 - 67 = 206 K

Therefore, the entropy change when 1 mol of HBr condenses is:

= [tex]\frac{-19.27 kJ/mol}{206 K} =0.0935 kJ/mol-K = -93.5 J/mol-K[/tex]

Thus for the given 1.80 mol of HBr the entropy change would be:

[tex]= 1.80 mol * -93.5 J/mol- K = -168.3 J/mol-K[/tex]

adioabiola

The entropy change at atmospheric pressure for the condensation of 1.80 moles of HBr is;

1.8 × -93.5 J/mol K = -168.3 J/mol K

The entropy of a system is simply defined as the degree of disorderliness in the system.

Mathematically, the change in entropy (molar) for a process is calculated as follows:

  • Delta S = Delta H/Temperature

The change in enthalpy, Delta H when HBr condenses, otherwise known as the Enthalpy of condensation is; -19.27 kJ/mol.

  • HBr condenses at -67°C

  • T = 273 - 67 = 206 K

  • Delta S = -19.27/206

  • Delta S = -0.0935kJ/mol K

  • Delta S = -93.5 J/mol K.

The entropy change calculated above is that of 1 mole of Hydrogen bromide.

Consequently, the entropy change for 1.80 moles of HBr is;

  • 1.8 × -93.5 J/mol K = -168.3 J/mol K

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