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    without doing any calculations, determine the sign of δssys for each of the following chemical reactions. drag the appropriate items to their respective bins.

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    Without doing any calculations, determine the sign of AS for each of the following chemical reactions. sys Drag the appropriate items to their respective bins. Reset Help As greater than 0 AS smaller than 0 HCI (g) + NH3 (g) NH,CI (s) PC4(1) + Ch(g)→PCI, (s) CHa(g) + 202(g)→ CO2(g) + 2H2O(1) 2H30+ (aq) + CO, (aq) (DOʻHE + (8) 0

    Without doing any calculations, determine the sign of AS for each of the following chemical reactions. sys Drag the appropriate items to their respective bins. Reset Help As greater than 0 AS smaller than 0 HCI (g) + NH3 (g) NH,CI (s) PC4(1) + Ch(g)→PCI, (s) CHa(g) + 202(g)→ CO2(g) + 2H2O(1) 2H30+ (aq) + CO, (aq) (DOʻHE + (8) 0

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    Transcribed Image Text:Without doing any calculations, determine the sign of AS for each of the following chemical reactions. sys Drag the appropriate items to their respective bins. Reset Help As greater than 0 AS smaller than 0 HCI (g) + NH3 (g) NH,CI (s) PC4(1) + Ch(g)→PCI, (s) CHa(g) + 202(g)→ CO2(g) + 2H2O(1) 2H30+ (aq) + CO, (aq) (DOʻHE + (8) 0

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    Determining the sign of ΔSsys for each given chemical reaction, the following can be obtained without calculations:1. ΔSsys > 0.2. ΔSsys < 03. ΔSsys < …

    08/11/2020 Chemistry College

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    Without doing any calculations, determine the sign of ΔSsys for each of the following chemical reactions. Drag the appropriate items to their respective bins.

    1. 2H30' (aq) + CO23- (aq) - CO2(g) +3H2O(1)

    2. CH4(g) + 202,(g) - CO2(g) + 2H2O(l)

    3. Mg (s) + Cl2(g) - MgCǐ2(s)

    4. SO3(g) + H2O(I) - H2SO4(I)

    A. ΔSsys greater than

    B. ΔSsys smaller than

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    Expert-Verified Answer

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    Determining the sign of ΔSsys for each given chemical reaction, the following can be obtained without calculations:

    1. ΔSsys > 0. 2. ΔSsys < 0 3. ΔSsys < 0 4. ΔSsys < 0 Recall:

    The states of the reactants and the products in a chemical reaction determines the sign of ΔSsys of the reaction.

    The entropy of gasses is greater than the entropy of liquid and solids.

    The entropy of solids is less than the entropy of liquid and gasses.

    Gasses have the highest entropy, while solids have the least.

    Thus:

    In the first chemical reaction, 1 mole of gas is produced, therefore: ΔSsys > 0.

    In the second chemical reaction, 3 moles of gasses gives a products of 1 mole of gas, therefore: ΔSsys < 0.

    In the third chemical reaction, 1 mole of gas gives 1 mole of solid as product, therefore: ΔSsys < 0.

    In the fourth chemical reaction, 1 mole of gas gives 1 mole of liquid as product, therefore: ΔSsys < 0.

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    Answer

    3 people found it helpful

    Answer:

    Answers are in the explanation.

    Explanation:

    In a chemical reaction we can determine the sign of ΔSsys based on the states of products and reactants knowing that:

    Entropy of gases >>> entropy of liquid > entropy of solids.

    The entropy of solids is lower than entropy of liquids that is lower than entropy of gases.

    In the reactions:

    1. 2H₃O⁺(aq) + CO₃²⁻(aq) → CO₂(g) +3H₂O(l)

    As 1 gas is produced, entropy of products is higher than entropy of reactants. That means  ΔSsys > 0 (That because ΔSsys is ΔSProducts - ΔSReactants)

    2. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

    3 moles of gas are converted in 1 mole of gas in products. Entropy of reactants is higher than entropy of products, ΔSsys < 0.

    3. Mg(s) + Cl₂(g) → MgCǐ₂(s)

    You have 1 mole of gas in reactants and 1 mole of solid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.

    4. SO₃(g) + H₂O(I) → H₂SO₄(I)

    1 mole of gas in reactants, a liquid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.

    5.0 (2 votes)

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