[vdemas]

Here's the question:
Use the information in Table 2.5 in Atkins (8th ed.) to predict the standard reaction enthalpy of  2 H₂(g) + O₂(g) = 2 H₂O(l) at 373K from its value at 273K.

Here's what I've done :

The standard reaction enthalpy at 273K is = [ (2)(-241 882) ] - [(2)(0) + (1)(0) ]
                                                                   = -483.63kJ/mol

Std rxn enthalpy at 373K = (-483.63kJ/mol) + [(2)(89.1) - (2)(28.824) + (1)(29.355)] x 75
                                              = -483.63kJ/mol + ( 178.2 - 87.003)(75)
                                              = -483.63kJ/mol + (91.197J/K.mol)(75K)
                                              = -483.63kJ/mol + 6.8397kJ/mol
                                              = -477kJ/mol

The values i underlined are the std molar heat capacities at constant pressure.
I am sure I did everything correct but the answer in the book is -566.96kJ/mol.
Can somebody please tell me where I went wrong.
Thanx in advance.

[Hunt]

First the temp is 298 K not 273K. you calculated delta T = 75 K based on that. Apart from the writing errors, your calculation is right but you used the wrong value of the standard enthalpy of formation of liquid ( not gas ) water . See page 997 atkins : Delta H_f = -285.83kJ/mol

You used the value for H2O gas.

[vdemas]

Thanx,I only realized now that I used the wrong values.
I checked out some other methods of doing it and one that really caught my eye was this :
There are 3 steps and the sum of these steps is the correct answer.

Step 1 : Cooling the reactants from 100⁰ celsius to 25^o celsius


H[ H₂O (g) ] = n C_p0 detlaT
                                      = (2) (28.824) (-75)
                                      = -4.3235kJ


H[ O₂ (g) ] = n C_p⁰ deltaT
                                    = (1) (29.355) (-75)
                                    = -2.2016kJ

The sum of these 2 values are : -6.5252kJ

Step 2 : Calculating the reaction enthalpy at 25⁰ celsius :


H = H (Products)- H (Reactants)
   = [(2) (-285.83)] - [ (2) (0) + (1) (0) ]
   = -571.66kJ


Step 3 : Warming the products from 25⁰ celsius to 100⁰ celsius :

H [ H₂O (g) ] = n C_p⁰ deltaT
                                       = (2) (33.58) (75)
                                       = 5.037kJ

The sum of all 3 steps = 5.037 + -571.66 + - 6.5252
                                    =-573.15kJ

Is this also away of doing it or should I stick to the previous method?

[Hunt]

Same method as before... try to give it a deeper look. They just divided it into steps. Add all the steps up and u'll get the same numerical equation in your 1st post. If it makes sense to you this way, then you can follow this procedure. You used the wrong Cp value though.

H [ H2O (g) ] = n Cp0 deltaT
                                       = (2) (33.58) (75)
                                       = 5.037kJ

It should be :  [(2)(89.1)](75) as you posted previously.

The answer would be exactly the same either way.

[vdemas]

Thank you very much, I got the correct answer.