# Comprehensive HW 2 Key

## 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

## 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.

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

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