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Extra Credit 14

  • Page ID
    96896
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    Q16.4.2

    Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.

    1. \(\ce{I2}(s)+\ce{Cl2}(g)⟶\ce{2ICl}(g) \hspace{20px} ΔG°=\mathrm{−10.88\: kJ}\)
    2. \(\ce{H2}(g)+\ce{I2}(s)⟶\ce{2HI}(g) \hspace{20px} ΔG°=\mathrm{3.4\: kJ}\)
    3. \(\ce{CS2}(g)+\ce{3Cl2}(g)⟶\ce{CCl4}(g)+\ce{S2Cl2}(g) \hspace{20px} ΔG°=\mathrm{−39\: kJ}\)
    4. \(\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g) \hspace{20px} ΔG°=\mathrm{−141.82\: kJ}\)
    5. \(\ce{CS2}(g)⟶\ce{CS2}(l) \hspace{20px} ΔG°=\mathrm{−1.88\: kJ}\)

    Q16.4.2 Solution

    a. We know that ΔG° =-RTln(K) and know that ΔG°=−10.88kJ, R is 8.314J/(K×mol) and it is at 298K( 25 °C)

    so plug in the the number in the formula get,

    -10.88kJ=-10880J=-8.314J/(K×mol)×298K×ln(K)

    Then compute get that K=80.75

    b. We know that ΔG° =-RTln(K) and know that ΔG°=3.4kJ, R is 8.314J/(K×mol) and it is at 298K( 25 °C)

    so plug in the the number in the formula get,

    3.4kJ=3400J=-8.314J/(K×mol)×298K×ln(K)

    Then compute get that K=0.254

    c. We know that ΔG° =-RTln(K) and know that ΔG°=−39kJ, R is 8.314J/(K×mol) and it is at 298K( 25 °C)

    so plug in the the number in the formula get,

    -39kJ=-39000J=-8.314J/(K×mol)×298K×ln(K)

    Then compute get that K=6.86× 106

    d. We know that ΔG° =-RTln(K) and know that ΔG°=−141.82kJ, R is 8.314J/(K×mol) and it is at 298K( 25 °C)

    so plug in the the number in the formula get,

    -141.82kJ=-141820J=-8.314J/(K×mol)×298K×ln(K)

    Then compute get that K=7.239× 1024

    e. We know that ΔG° =-RTln(K) and know that ΔG°=−1.88kJ, R is 8.314J/(K×mol) and it is at 298K( 25 °C)

    so plug in the the number in the formula get,

    -1.88kJ=-1880J=-8.314J/(K×mol)×298K×ln(K)

    Then compute get that K=2.136

    Q16.4.2 Answer:

    a. ΔG° =-RTln(K) -10.88kJ=-10880J=-8.314J/(K×mol)×298K×ln(K) K=80.75

    b. ΔG° =-RTln(K) 3.4kJ=3400J=-8.314J/(K×mol)×298K×ln(K) K=0.254

    c. ΔG° =-RTln(K) -39kJ=-39000J=-8.314J/(K×mol)×298K×ln(K) K=6.86× 106

    d. ΔG° =-RTln(K) -141.82kJ=-141820J=-8.314J/(K×mol)×298K×ln(K) K=7.239× 1024

    e. ΔG° =-RTln(K) -1.88kJ=-1880J=-8.314J/(K×mol)×298K×ln(K) K=2.136

    Q5.4.22

    Calculate \(ΔH^\circ_{298}\) for the process \(\ce{Zn}(s)+\ce{S}(s)+\ce{2O2}(g)⟶\ce{ZnSO4}(s)\)

    from the following information:

    \(\ce{Zn}(s)+\ce{S}(s)⟶\ce{ZnS}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−206.0\:kJ}\)

    \(\ce{ZnS}(s)+\ce{2O2}(g)⟶\ce{ZnSO4}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−776.8\:kJ}\)

    Q5.4.22 Solution

    First plus the two chemical together we get

    \(\ce{Zn}(s)+\ce{S}(s)+\ce{ZnS}(s)+\ce{2O2}(g)⟶\ce{ZnS}(s)+\ce{ZnSO4}(s)\hspace{20px}\)

    Then can cancel out the same element appear on bother side of equation, which is ZnS, we get

    \(\ce{Zn}(s)+\ce{S}(s)+\ce{2O2}(g)⟶\ce{ZnSO4}(s)\)

    Since we add up the two chemical equation, so the \(ΔH^\circ_{298}\) for the process is just add up the two ΔH of the chemical

    \(ΔH^\circ_{298}\)= -206.0kJ+(-776.8kJ)

    \(ΔH^\circ_{298}\)=-982.8kJ

    Q5.4.22 Answer:

    \(ΔH^\circ_{298}\)= -206.0kJ+(-776.8kJ)=-982.8kJ

    Q10.5.8

    Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances:

    1. CaCl2
    2. SiC
    3. N2
    4. Fe
    5. C (graphite)
    6. CH3CH2CH2CH3
    7. HCl
    8. NH4NO3
    9. K3PO4

    Q10.5.8 Solution

    1. CaCl2 Ca2++Cl- ⟶ CaCl2 so it is ionic
    2. SiC SiC will bond with other SiC so it is network covalent
    3. N2 Nitrogen and nitrogen is covalent bond so it is molecular
    4. Fe Iron is a metal so metallic
    5. C (graphite) Carbon connect with covalent bonds so it is network covalent
    6. CH3CH2CH2CH3 The compound is covalently bonded by the elements together so it is molecular
    7. HCl HCl is hydrogen and chlorine are covalently bound so it is a molecular
    8. NH4NO3 NH4++NO3- ⟶ NH4NO3 so it is ionic
    9. K3PO4 K++PO43- ⟶ K3PO4 so it is ionic

    Q10.5.8 Answer:

    a. ionic b. network covalent c. molecular d. metallic e. molecular f. molecular g. molecular h. ionic i. ionic

    Q13.3.17

    How can the pressure of water vapor be increased in the following equilibrium?

    \[\ce{H2O}(l)\rightleftharpoons\ce{H2O}(g) \hspace{20px} ΔH=\ce{41\:kJ}\]

    Q12.3.17 Solution

    Since the process is endothermic so increased the temperature of the system will made the process go toward the side without heat. Therefore, if we increase the temperature then there will be more water vapor produce and cause the pressure go up.

    Q13.3.17 Answer:

    Increase the temperature.

    Q14.3.8

    Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.

    \(\underset{\large\ce{BB}}{\ce{Mg(OH)2}(s)}+\underset{\large\ce{BA}}{\ce{HCl}(aq)}⟶\underset{\large\ce{CB}}{\ce{Mg^2+}(aq)}+\underset{\large\ce{CA}}{\ce{2Cl-}(aq)}+\underset{\:}{\ce{2H2O}(l)}\)

    Q14.3.8 Solution

    I think the answer is already give to us.

    \(\underset{\large\ce{BB}}{\ce{Mg(OH)2}(s)}+\underset{\large\ce{BA}}{\ce{HCl}(aq)}⟶\underset{\large\ce{CB}}{\ce{Mg^2+}(aq)}+\underset{\large\ce{CA}}{\ce{2Cl-}(aq)}+\underset{\:}{\ce{2H2O}(l)}\)

    In the equation that Mg(OH)2 is the base and the conjugate acid is 2Cl-, and the HCl is the base and the conjugate acid is Mg2+. Water is just the product.

    Since the MgCl2 is a salt so the equation will be like that

    \({\ce{Mg(OH)2}(s)}+{\ce{HCl}(aq)}⟶{\ce{MgCl2}(aq)}+{\ce{2H2O}(l)}\)

    Q14.3.8 Answer:

    Mg(OH)2(s)+HCl(aq)⟶MgCl2(aq)+H2O(l)

    acid conjugate base
    H2O H3O+
    HCl Cl-


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