Question

Determine the entropy change when 1.80 mol of HBr(g) condenses at atmospheric pressure?

Answer 1

Answer:

Entropy change = -168.3 J/mol-K

Explanation:

The entropy change is expressed as:

$DeltaS = \frac{DeltaH}{T}$

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:

= $\frac{-19.27 kJ/mol}{206 K} =0.0935 kJ/mol-K = -93.5 J/mol-K$

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

$= 1.80 mol * -93.5 J/mol- K = -168.3 J/mol-K$

Answer 2

Hvap, Tvap, Mr are constants
Hvap = 443.38 kJ/kg = 196417.34 J/mol
Tvap = -85.1 °C
Mr = 36.5 g/mol
n = 1.80 mol

ΔS = n*Hvap/Tvap
ΔS = 1.80 * 196417.34 / (273-85.1)
ΔS = 1881.59 J/K = 1.882 kJ/K