Ask Question
8 June, 12:31

Calculate the standard entropy change for the following reactions at 25°C.

(a) C2H5OH (l) + 3 O2 (g) 2 CO2 (g) + 3 H2O (g)

(b) CS2 (l) + 3 O2 (g) CO2 (g) + 2 SO2 (g)

(c) 2 C6H6 (l) + 15 O2 (g) 12 CO2 (g) + 6 H2O (g)

+1
Answers (1)
  1. 8 June, 14:19
    0
    (a) ΔSº = 216.10 J/K

    (b) ΔSº = - 56.4 J/K

    (c) ΔSº = 273.8 J/K

    Explanation:

    We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

    First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.

    (a) C2H5OH (l) + 3 O2 (g) ⇒ 2 CO2 (g) + 3 H2O (g)

    Sº 159.9 205.2 213.8 188.8

    (J/Kmol)

    ΔSº = [ 2 (213.8) + 3 (188.8) ] - [ 159.9 + 3 (205.) ] J/K

    ΔSº = 216.10 J/K

    (b) CS2 (l) + 3 O2 (g) ⇒ CO2 (g) + 2 SO2 (g)

    Sº 151.0 205.2 213.8 248.2

    (J/Kmol)

    ΔSº = [ 213.8 + 2 (248.2) ] - [ 151.0 + 3 (205.2) ] J/K = - 56.4 J/K

    (c) 2 C6H6 (l) + 15 O2 (g) 12 CO2 (g) + 6 H2O (g)

    Sº 173.3 205.2 213.8 188.8

    (J/Kmol)

    ΔSº = [ 12 (213.8) + 6 (188.8) ] - [ 2 (173.3) + 15 (205.2) ] = 273.8 J/K

    Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.

    Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Calculate the standard entropy change for the following reactions at 25°C. (a) C2H5OH (l) + 3 O2 (g) 2 CO2 (g) + 3 H2O (g) (b) CS2 (l) + 3 ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers