Adding and subtracting thermalchemcial equations help needed

[chemistry4all]

Homework Statement

From the enthalpies of reaction for equations A) and B), calculate the heat of reaction for equation C).

A) 2H2 (g) + O2 (g) ----> 2H2O2 (g) DH= -483.6 kJ
B) 3O2 (g) ----> 2O3 (g) DH= +284.6 kJ
c) 3 H2 (g) + O3 (g) -----> 3H2O (g) DH= ? kJ

Homework Equations

NONe

The Attempt at a Solution

My attempt is as follows:

I multiplied equation A) by 3 first including the DH as well and got and flipped equation be so the 2O3 can cancel out.

A) 6H2 (g) + 3O2 (g) ----> 6H2O (g) DH= -1450.8 kJ
B) 2O3 (g) ------> 3O2 DH= -284.6 kJ

adding A) + B) I got

6H2 (g) + 2O3 (g) -------> 6H2O (g) DH= -1699.4 kJ


since i need to get to c) 3 H2 (g) + O3 (g) -----> 3H2O (g) all I did is divide everything by 2 including the DH = -1699.4 kJ

the DH came out to for equation C) as -849.7 kJ and this is not correct because the enthalpy for H2O (g) is -241.82 and since there are 3 H2O it should be -725.46 which I can't seem to get.

Can someone please help where I have made a mistake? Thanks

 

[ehild]

the DH came out to for equation C) as -849.7 kJ and this is not correct because the enthalpy for H2O (g) is -241.82 and since there are 3 H2O it should be -725.46 which I can't seem to get.

What does it mean that the enthalpy of H2O(g) is -241.82? Does not it refer to a reaction from the elementary constituents, H2 and O2?

ehild

 

[chemistry4all]

the -241 is from the data table of standard formations of enthalpy but in this problem it does not want me look it up it wants me to calculate it using the two equations. The answer should be double the -241 but its not coming out that way. I think I did something wrong and I have tried everything to see what I did wrong. Can't seem to figure it out.


Anyone please help?

 

[Onamor]

as ehild said, the 242 you've quoted is probably from standard constituents.
For H2O, this is probably H2 and 1/2 O2, not O3.
but C) asks for H2 + O3.

 

[chemistry4all]

I know this is wrong because the first equation if you look at the standard formation of enthalpies chart is correct and so is the number for O2 ----> O3 is correct with the proper coeffients that are already in front of it. So mathematically the equation A and B should add up be able to get to equation C right?

As far as I know these are the rules:

1) All the reactants in the overall equation must appear on the left side.
2) All the products in the overall equation must appear on the right side.
3) All reaction intermediates (those not present in the overall equation) must appear on both sides equally so they are canceled out.
4) You sum up the individual equations (just like a standard addition problem) to make the overall equation.
5) A reaction written in reverse of the direction given in the problem must have the sign of its enthalpy changed.
6) A reaction multiplied by a coefficient in order to balance the overall equation must have its enthalpy multiplied by the same coefficient.
I did exactly the rules above unless I miss something.

 

[Redbelly98]

The Attempt at a Solution

My attempt is as follows:

I multiplied equation A) by 3 first including the DH as well and got and flipped equation be so the 2O3 can cancel out.

A) 6H2 (g) + 3O2 (g) ----> 6H2O (g) DH= -1450.8 kJ
B) 2O3 (g) ------> 3O2 DH= -284.6 kJ

adding A) + B) I got

6H2 (g) + 2O3 (g) -------> 6H2O (g) DH= -1699.4 kJ

Check your addition in this last step. Otherwise, your procedure is correct.

Your argument about getting 3x the enthalpy of H2O is not valid, as others have pointed out.