Question

Determine the entropy change when 2.40 mol hi(g) condenses at atmospheric pressure.

Answer 1

The entropy change for 2.40 moles of HI when it condenses at atmospheric pressure is -199.968 J/mol K

What is the entropy change?

Entropy change is defined as the measure of change of disorder or randomness in a thermodynamic system.

Mathematically, it is given as

$\Delta S = \frac{\Delta H}{T}$

where ∆S = entropy change

∆H = change in enthalpy

T = temperature

For HI,

Enthalpy of vaporization of HI, $\Delta H_{vap}$ = 19.8 kJ/mol

We also know

Enthalpy of vaporization of HI, $\Delta H_{vap}$ = - Enthalpy of condensation of HI, $\Delta H_{cond}$

$\Delta H_{cond}$ = -19.8 kJ/mol

HI condenses at -35.36°C

Temperature, T = -35.36°C + 273 = 237.64 K

Substitute in the entropy change formula,

$\Delta S = \frac{-19.8}{237.64}$

$\Delta S$ = -0.08332 kJ/mol K = -83.32 J/mol K

1 mole of HI has entropy change = -83.32 J/mol K

2.4  moles of HI have entropy change = 2.4 × (-83.32) = -199.968 J/mol K

Thus, The entropy change for 2.40 moles of HI when it condenses at atmospheric pressure is -199.968 J/mol K

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