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Comprehensive HW 3 Key

  • Page ID
    206254
  • Q

    FALSE. \(\Delta G\) has no bearing on the kinetics of the reaction. Ever.

     

    Q

    This set of questions uses the equation

    \(\Delta G = \Delta H -T \Delta S \)

    a. Unknown it depends on the magnitude of the Enthalpy and Entropy.

    b. Spontaneous. \( \Delta G < 0 \)

    c. Stationary. \(\Delta G > 0 \)

    d. Unknown it depends on the magnitude of the Enthalpy and Entropy.

     

    Q

    a. This is a comparison between distributions (c) and (e).

    \(S = k ln(W) \)

    \(\Delta S = k ln (1) - k ln(6) = k ln (1/6) \)

    \( \Delta S = -1.7 k = -2.4 \times 10^{-23} J/K \)

    b. A comparison of distributions (b) and (d)

    \(\Delta S = 0 \)

    Both systems have the same weight so there is no change in entropy.

     

    Q

    So this problem has roots in the Partition Function and Statistical Mechanics. I will not attempt a numerical answer because you've not learned that yet. Wait until PChem Yay!

    But simply the more degrees of freedom and motion a molecule has the larger the number of microstates available to it the larger the entropy.

    a. \( HBrO_4 (g) > HBr (g) >H_2(g) \)

    b. \( H_2O (g) > H_2O (l) > H_2O (s) \)

    c. \(P_4(g) > Cl_2 (g) > He (g) \)

     

    Q

    Same idea as the last problem. Heavier atoms have a larger density of states for translations. See the particle in the box Energy equation.

    \( I_2 (g) > Br_2 (g) > Cl_2 (g) \)

     

    Q

    More...

    a. \(C_3H_7OH (l) \) Heavier and more atoms so more vibrational entropy.

    b. \(C_2H_5OH (g) \) Gas!! and heavier and more atoms so more vibrational entropy.

    c.\(2H (g) \) More atoms, More possible configurations.

     

    Q

    Explosion!

    \(C_6H_6(g) + 7.5O_2(g) \rightarrow 6CO_2 (g) + 3H_2O \)

    The entropy should increase becasue there are more mols of gas on the products side than on the reactants side.

     

    Q

    \(\Delta G \) is the Gibbs free energy at whatever temperature, pressure and # of mols.

    \(\Delta G^o \) is the value for 1 mol of the substance in it's standard state. The temperature and pressure may be varied as required. For example \(\Delta G^o_{f, CO_2(s)} \) only applies at -78.5 C.

    \( \Delta G^o_{298} \) is the value for 1 mol of the substance at 298 K and 1 atm.

     

    Q

    \(\Delta G = \Delta H -T \Delta S \)

    \( \Delta H_{reaction} = \Delta H_{f,NH_3} + \Delta H_{f,HCl} - \Delta H_{f,NH_4Cl} \)

    \( \Delta H_{reaction} = -45 + -92.3 - -314 = 177 kJ/mol \)

    \( \Delta S_{reaction} = S_{f,NH_3} + S_{f,HCl} - S_{f,NH_4Cl} \)

    \( \Delta S_{reaction} = 192 + 186 - 94 =284 J/m \)

    \(\Delta G = 177,000 - 300 \times 284 = 92 kJ \)

    This reaction is stationary. It is entropically favorable and enthalpically unfavorable.

     

    Q

    Burning Methane

    \( CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + FIRE! \)

    \( \Delta G_{reaction} = \Delta G_{f,CO_2} + 2 \Delta G_{f,H_2O} - \Delta H_{f,CH_4} - 2 \Delta H_{f,O_2} \)

    \( \Delta G_{reaction} = -394 + 2 (-228) - -75 - 0 = - 775 kJ/mol \)

     

    Q

    a. Decreases moving from a liquid to a solid.

    b. Increases moving from a liquid to a gas

    c. Decreases, more mols of gas in the reactants.

     

    Q

    Its an add reactions problem....

    \(\ce{N}(g)+\ce{O}(g)⟶\ce{NO}(g) \hspace{20px} ΔS^\circ_{298}=\,?\)

    \(\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g) \hspace{20px} ΔS^\circ_{298}=\mathrm{24.8\: J/K}\)

    \(\ce{N2}(g)⟶\ce{2N}(g) \hspace{20px} ΔS^\circ_{298}=\mathrm{115.0\: J/K}\)

    \(\ce{O2}(g)⟶\ce{2O}(g) \hspace{20px} ΔS^\circ_{298}=\mathrm{117.0\: J/K}\)

    Hess' Law

    \( \Delta S = (\Delta S_{rxn1} - \Delta S_{rxn2} - \Delta S_{rxn3}) \times 1/2 \)

    \( \Delta S = (24 - 115 - 117) \times 1/2 = 104 \)

     

    Q

    I think it wants to find \(\Delta G\)

    \(\Delta G = \Delta H -T \Delta S \)

    \(\Delta G = 27,950 J/mol - 800K \times 23 J/molK = 9550 \)

    \(\Delta G = 27,950 J/mol - 1000K \times 23 J/molK = 4950 \)

    So neither of these is hot enough to melt salt. What about 1000 C?

    \(\Delta G = 27,950 J/mol - 1300K \times 23 J/molK = -1950 J \)

    This is much higher than the actual melting point of NaCl at 800 C. Whats wrong?

    We assumed that the standard entropy at 25 C applied at 1000 C. It's obviously not true so be careful in using any thermodynamic value without considering if it is valid at the temperature in your reaction.

     

    Q

    Since I love NASA I know the answer is Hydrazine. Its what they used in the Apollo missions and the space shuttle. And they used Hydrogen Peroxide on the Mercury missions (had to look that up).

    To "calulate" the answer we are looking for reactions that have \(\Delta G < 0 \) and \(\Delta H < 0 \).

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