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Chemistry LibreTexts

Comprehensive HW 2 Key

  • Page ID
    206253
  • Q

    1. Exothermic - Uh. It heats up

    2. Exothermic - Why you sweat to cool off when you exercise.

    3. Endothermic - Same reaction as the Thermochemistry lab.

     

    Q

    This is a simple calculation of Q.

    \(Q=mC \Delta T = 150 g \times 4.18 J/g \times (25.8-22.4) = 2131 J \)

    The temperature is rising - reaction is exothermic.

     

    Q

    Another Q problem except this time you need to include the calorimeter

    \(Q=mC \Delta T + C_{cal} \Delta T = 675 g \times 4.18 J/gK \times (26.9-23.4C) + 534 J/K \times (26.9-23.4C) = 11,700 J \)

     

    Q

    a. 100 J of energy is added to a system that does 40 J of external work.

    \( Q=100 J\) and \(W = 40 J\)

    \(\Delta U = Q + W = 100 - 40 = 60 J \)

    b. 75 J of energy is removed from a system that has 50 J of work done on it.

    \(Q= -75 J\) and \( W= -50 J\)

    \( \Delta U= Q + W = -75-(-50) = -25 J \)

    The internal energy has decreased by 25 J

    c.150 J of energy is added to a system that has 75 J of work done of it?

    \(Q= 150 J \) and \(W= -75 J\)

    \( \Delta U = Q + W = 150-(-75) = 225 J\)

     

    Q

    More Q problems with no Calorimeter heat capacity

    \(Q=mC \Delta T = 104.7 g \times 4.2 J/g \times -3.1K = -1360 J \)

     

    Q

    This is is Hess' Law equation.

    \(\Delta H_{reaction} = \sum \Delta H_{Products} - \sum \Delta H_{Reactants} \)

    \( \Delta H_{reaction} = \Delta H_{f,Al_2O_3} + 2\Delta H_{f,Fe} - \Delta H_{f,Fe_2O_3} - 2\Delta H_{f,Al} \)

    \( \Delta H_{reaction} = -1670 kJ/mol + 0 - -826 kJ/mol - 0 = 840 kJ \)

    This is very exothermic. Hot enough to make molten Iron

    https://www.youtube.com/watch?v=a8XSmSdvEK4

     

    Q

    Smaller; more heat would be lost to the coffee cup and the environment and so ΔT for the water would be smaller and the calculated q would be smaller.

     

    Q

    \[ \dfrac{-1}{2} \times [N_2(g) + 3 H_2 (g) \rightarrow 2 NH_3 (g)] \tag{-1/2 x ΔH = -1/2(-82.6 kJ)}\]

    \[ -1 \times [C (s) + 2 H_2 (g) \rightarrow CH_4 (g)] \tag{-1 x ΔH = -1(-90.3 kJ)}\]

    \[ \dfrac{1}{2} \times [ H_2 (g) + 2 C (s) + N_2 (g) \rightarrow 2 HCN (g)] \tag{1/2 x ΔH = 1/2 (293.3 kJ)}\]

    \[\Delta{H}=\sum_i \Delta{H_i} = -\dfrac{-1}{2}(-82.6 \; kJ) + -1(-90.3\; kJ) + \dfrac{1}{2} (293.3\; kJ) = 278.25 \; kJ\]

     

    Q

    Hess' Law

    \( \Delta H_{reaction} = \Delta H_{f,MgO} + \Delta H_{f,CO_2} - \Delta H_{f,MgCO_3} \)

    \(\Delta H= (-601.7)kJ + (-393.5)kJ - (-1096)kJ = 101 kJ \)

     

    Q

    Per Hess’ Law, the enthalpy of the reaction is equal to the heat of formation of the products minus reactants.

    \(\Delta H_{rxn} = 6*(-393.5) + 6*(-285.8) - (-1275.0 + 6*0) = -2801 kJ/mol. \)

     

    Q

    a.

    \( \Delta H_{reaction} = \Delta H_{f,CO} + \Delta H_{f,H_2} - \Delta H_{f,Coke} - \Delta H_{f,H_2O} \)

    \( \Delta H_{reaction} = -110 + 0 - 0 - - 241 = 131 kJ \)

    b.

    \( \Delta H_{reaction} = \Delta H_{f,CH_3OH} - 2\Delta H_{f,H_2} - \Delta H_{f,CO} \)

    \( \Delta H_{reaction} = -201 - 0 - -110 = -91 kJ \)

    The condensation of methanol vapor to methanol liquid is just another reaction. Don't treat it any different.

    \( \Delta H_{reaction} = \Delta H_{f,CH_3OH (l)} - \Delta H_{f,CH_3OH (g)} \)

    \( \Delta H_{reaction} = -238 - -201 = -37 kJ \)

    c. Another Hess' Law reaction But this one explodes ;) Remember to balance the reaction.

    \( \Delta H_{reaction} = \Delta H_{f,CO_2} + 2\Delta H_{f,H_2O} - \Delta H_{f,CH_3OH} - 1.5\Delta H_{f,O_2} \)

    \( \Delta H_{reaction} = -393 + 2 \times -241 - -201 - 0 = 674 kJ \)

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