Calculating Energy Changes in Chemical Reactions

[jesusismagic28]

Help on a few homework problems?

1). Use the bond dissociation energies in Table 7.1 to calculate an approximate H° for the industrial synthesis of isopropyl alcohol (rubbing alcohol) by reaction of water with propene.
I'm given: CH3CH=CH2+H2O -> CH3CHCH3 [with an OH bond coming from the second carbon atom]

2). Assume that the kinetic energy of a 2300 kg car moving at 115 km/h could be converted entirely into heat. What amount of water could be heated from 20.°C to 45°C by the car's energy?

3). The addition of H2 to C=C double bonds is an important reaction used in the preparation of margarine from vegetable oils. If 50.0 mL of H2 and 50.0 mL of ethylene (C2H4) are allowed to react at 1.5 atm, the product ethane (C2H6) has a volume of 50.0 mL. Calculate the amount of PV work done, and tell the direction of the energy flow.
C2H4(g) + H2(g) C2H6(g)

4). Assume that a particular reaction evolves 227 kJ of heat and that 51 kJ of PV work is gained by the system. What are the values of H and E for the system? What are the values of H and E for the surroundings?

5). Instant cold packs used to treat athletic injuries contain solid NH4NO3 and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves, lowering the temperature because of the following endothermic reaction.
NH4NO3(s) + H2O(l) NH4NO3(aq) H = +25.7 kJ

What is the final temperature in a squeezed cold pack that contains 50.0 g of NH4NO3 dissolved in 115 mL of water? Assume a specific heat of 4.18 J/(g·°C) for the solution, an initial temperature of 25.0°C, and no heat transfer between the cold pack and the environment.

6). Calculate H°f for benzene, C6H6, from the following data.
2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(l) H° = -6534 kJ

H°f (H2O) = -285.8 kJ/mol
H°f (CO2) = -393.5 kJ/mol



Any help would be greatly appreciated. I've tried all of them several different ways and cannot figure it out.

 

[eli64]

a few?? problems

can you show us how you tried to solve these, we can help you better if we see what you've already done - as you tried them several different ways, just give us the ones that are your best guess

- include what equations you are using

 

[Blueshift5]

1) To calculate the approximate H° for the reaction, we need to use the bond dissociation energies for the bonds broken and formed in the reaction. In this case, we have one C=C bond and one O-H bond being broken, and one C-C bond and one O-H bond being formed. Using the bond dissociation energies from Table 7.1, we can calculate the total energy change:

ΔH° = [1(837 kJ/mol) + 1(467 kJ/mol)] - [1(413 kJ/mol) + 1(463 kJ/mol)]
= 841 kJ/mol

Therefore, the approximate H° for the reaction is 841 kJ/mol.

2) To calculate the amount of water that can be heated by the car's kinetic energy, we can use the equation Q = mcΔT, where Q is the heat energy, m is the mass of the water, c is the specific heat of water (4.18 J/(g·°C)), and ΔT is the change in temperature.

First, we need to convert the car's kinetic energy from km/h to m/s:
115 km/h = 31.94 m/s

Next, we can calculate the kinetic energy of the car:
KE = 1/2mv^2 = 1/2(2300 kg)(31.94 m/s)^2 = 1.17 x 10^6 J

Finally, we can use this energy to calculate the amount of water that can be heated:
Q = (1.17 x 10^6 J)(1 cal/4.18 J)(25°C) = 6.7 x 10^5 cal

Therefore, the car's kinetic energy can heat 6.7 x 10^5 cal of water, which is equivalent to 6.7 x 10^5/4.18 = 1.6 x 10^5 g = 160 kg of water.

3) To calculate the PV work done in the reaction, we can use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature.

First, we need to convert the volume from mL to L:
50.0 mL = 0.050