How do you calculate entropy?

1 Answer
Apr 17, 2014

Here are some formulas.

ΔS_"total" = ΔS_"univ" = ΔS_"sys" + ΔS_"surr" = q_"sys"/T_"sys" + q_"surr"/T_"surr"

where q is the heat and T is the Kelvin temperature.

Entropy Change for the System

ΔS_"sys" = ΔS_"rxn" = Sigma(n_pS_"products"^"o") – Sigma(n_rS_"reactants"^"o")

where n_p and n_r represent moles of products and reactions.

Entropy Change for the Surroundings

ΔS_"surr" = q_"surr"/T_"surr"

q_"surr" = -q_"sys"

ΔS_"surr" = q_"surr"/T_"surr" = -q_"sys"/T_"surr"

Example 1:

What is ΔS_"surr" at 300 K for the reaction

reactants → products; ΔH = 75 kJ

Solution:

ΔS_"sys" = q/T = (75 000" J")/(300" K") = 250 J/K

ΔS_"surr" = q_"surr"/T_"surr" = -(ΔH_"sys")/T_"surr" = -(75 000" J")/(300" K") = -250 J/K

Example 2:

What is ΔS_"rxn"^"o" for the following reaction?

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

The S^"o" values are NH₃ = 193 J·K⁻¹mol⁻¹; O₂ = 205 J·K⁻¹mol⁻¹; NO = 211 J·K⁻¹mol⁻¹;
H₂O = 189 J·K⁻¹mol⁻¹

ΔS_"sys" = Sigma(n_pS_"products"^"o") – Sigma(n_rS_"reactants"^"o")

ΔS_"rxn"^"o" = 4S_"NO"^"o" + 6S_"H₂O"^"o" – 4S_"NH₃" - 5S_"O₂"^"o"

ΔS_"rxn"^"o" = (4×211 + 6 × 189 - 4 × 193 + 5 × 205 ) J/K⁻¹ = 181 J•K⁻¹