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Calculating the entropy change of the system

  • 3 things to know
  • Components 01, 02 and 03

Overview

Recall questions on using standard entropies, multiplying by the stoichiometric numbers, and why elements do not have zero standard entropy.

What you need to know

Answer each one before you reveal it.

Your recallNot started
Numbers

How is the entropy change of the system found from standard entropies, why is each multiplied by its balancing number, and why are the standard entropies of elements not zero?

Key term

Why is the entropy change of the system for the vaporisation of water large and positive?

Key term

Why are standard entropies quoted at a stated temperature and pressure?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    How is the entropy change of the system calculated from standard entropy data?

    Model answer

    ΔsysS is the sum of the standard entropies of the products minus the sum of the standard entropies of the reactants.

  2. 2

    Why must each standard entropy be multiplied by the number in front of the formula in the equation?

    Model answer

    Entropy is an extensive quantity, so two moles of a substance contribute twice the entropy of one mole.

  3. 3

    What are the units of a standard entropy value?

    Model answer

    They are J K⁻¹ mol⁻¹.

  4. 4

    Why are the standard entropies of elements not zero, unlike their standard enthalpies of formation?

    Model answer

    Entropy is measured from a genuine zero at absolute zero, so any substance above 0 K has a real positive entropy.

  5. 5

    Using S(N₂) = 191.6, S(H₂) = 130.6 and S(NH₃) = 192.3 J K⁻¹ mol⁻¹, calculate ΔsysS for N₂(g) + 3H₂(g) → 2NH₃(g).

    Model answer

    ΔsysS = 2 × 192.3 − (191.6 + 3 × 130.6) = 384.6 − 583.4 = −199 J K⁻¹ mol⁻¹ to three significant figures.

  6. 6

    Using S(CaCO₃) = 92.9, S(CaO) = 39.7 and S(CO₂) = 213.6 J K⁻¹ mol⁻¹, calculate ΔsysS for CaCO₃(s) → CaO(s) + CO₂(g).

    Model answer

    ΔsysS = (39.7 + 213.6) − 92.9 = +160.4 J K⁻¹ mol⁻¹.

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