Thermochem question!?

Question

1. A 15.0 g piece of a metal at 100.oC (Cp = 0.4484 J/goC) is placed in 100. ml of water at 20.0oC. Determine the final temperature of the iron/water system.

also

From the following information, determine the enthalpy of formation of nitrogen dioxide from its elements.

N2 (g) + 2 O2 (g)  2NO2 (g)

Step #1: N2 (g) + O2 (g)  2NO ΔH = 180 kJ/mol

Step #2: NO2 (g)  ½ O2 (g) + NO (g) ΔH = 61 kJ/mol

any help is appreciated. solutions would be best!

Answer 1

1.  As you probably know, the heat lost by the metal will equal the heat gained by the water.  The heat lost by the metal will look like this:  

Heat Lost (in Joules) = (15.0g)(0.4484 J/(g*oC))(100 oC - x)  where x is the final temperature.

The heat gained by the water will look like this:

Heat Gained (Joules) = (100g)(4.184 J/(g*oC))(x-20 oC)

and again x is the final temperature.  

For the heat gained by water, you had to know 2 things about water that aren't given: 1) the density of water is 1.0000 g/mL; and 2) the specific capacity of water is 4.184 J/goC.  Since water is such an important and common molecule, you will be expected to know just about everything about it going into a test, meaning it probably won't be given.  Besides these values, memorize its molecular weight (18.02) and maybe even its (delta)H values (vaporization=334 J/g, melting=2.27 kJ/g)

Since the heat lost = heat gained, set the two equations equal:  (15)(0.4484)(100-x) = (100)(4.184)(x-20)

Solving for x should put you at around the 21 oC mark.

2.  The bulk of the work you have to do with this is get the right stoichiometry and the right reaction direction in Step #2.  Since NO2 is your final product, you need to reverse it:

0.5 O2 + NO -> NO2

And in doing so, your (delta)H will be inverted to -61 kJ/mol.  You can also see that your overall equation calls for 2 moles of NO2, meaning that everything will be doubled, including (delta)H:

O2 + 2NO -> 2NO2    (delta)H = -122 kJ/mol

When you add this equation to Step #1, you should find that it indeed equals the equation first given.  The enthalpy is then also the sum of the two equations, that is 58 kJ/mol.

Answer 2

1. There is 100C - 20C = 80C temperature difference between the iron and the water. Let T be the temperature rise of the water. Then 80 - T is the temperature drop of the iron. The heat lost by the iron has to equal the heat gained by the water. The heat capacity of water is 4.184 J/g-C.

15.0gFe x (80-T)degC x 0.4484J/g-C = 100gH2O x (T)degC x 4.184J/g-C

538 - 6.7T = 418.4T

538 = 425.1T

T = 1.26 degC

So the final temperature is 21.3C or so

2. N2 + O2 ===> 2NO  /\H = +180kJ

Reverse the second equation and multiply it by 2:

2NO + O2 ===> 2NO2  /\H = -122kJ

Add the equations and heats together, collecting terms:

N2 + 2O2 ===> 2NO2  /\H = +58kJ

Are you sure you have the right signs on these two heats of reaction?