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Calculate the entropy change for the reaction: HCl(g) + NH3(g) -> NH4Cl(s) Entropy data: HCl: 187

Posted on October 23, 2021 By Cocobelle 3 Comments on Calculate the entropy change for the reaction: HCl(g) + NH3(g) -> NH4Cl(s) Entropy data: HCl: 187

Calculate the entropy change for the reaction: HCl(g) + NH3(g) -> NH4Cl(s) Entropy data: HCl: 187 J/K mol NH3: 193 J/K mol NH4Cl: 94.6 J/K mol

Chemistry

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Comments (3) on “Calculate the entropy change for the reaction: HCl(g) + NH3(g) -> NH4Cl(s) Entropy data: HCl: 187”

  1. Expert says:
    October 23, 2021 at 7:45 am

    a is the answer to this question

    explanation:

    Reply
  2. Expert says:
    October 23, 2021 at 2:51 pm

    since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

    Reply
  3. gena75 says:
    October 24, 2021 at 12:20 am

    -285.4 J/K

    Explanation:

    Let's consider the following balanced equation.

    HCl(g) + NH₃(g) ⇒ NH₄Cl(s)

    We can calculate the standard entropy change for the reaction (ΔS°r) using the following expression.

    ΔS°r = 1 mol × S°(NH₄Cl(s)) - 1 mol × S°(HCl(g)) - 1 mol × S°(NH₃(g))

    ΔS°r = 1 mol × 94.6 J/K.mol - 1 mol × 187 J/K.mol - 1 mol × 193 J/K.mol

    ΔS°r = -285.4 J/K

    Reply

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