Hess' Law Calculation for Reaction A: \Delta H = 121 kJ | Reference Table Used

[DB]

I don't have the answer to this, and it's a big question so I'd like to know if I got it right. Thanks

Determine the \Delta H value for reaction A:

A)3CH_4 \rightarrow C_3H_8 + 2H_2

\Delta H=?

Okay so I was given a reference table and used these two equations for Hess' Law:

B)C+2H_2 \rightarrow CH_4

\Delta H= -74.9

C)3C+4H_2 \rightarrow C_3H_8

\Delta H=-103.7

So I knew I the sum of B and C would have to give me A, so in order for the values to cancel properly I had to multiply B by three:

B)3(C+2H_2 \rightarrow CH_4)

3(\Delta H=-74.9)

gives me:

B')3C+6H_2 \rightarrow 3CH_4

\Delta H=-224.7

Then, since propane is on the left side of equation A, I would have to reverse B':

3CH_4 \rightarrow 3C + 6H_2

\Delta H=+224.7

Now I can add them:

3CH_4 +3C - 3C \rightarrow C_3H_8 + 6H_2 - 4H_2

gives:

3CH_4 \rightarrow C_3H_8 + 2H_2

which is equation A.

So now I must add the change in enthalpy:

\Delta H=+224.7+\Delta H=-103.7

\Delta H=121 kJ for reaction A

is that right?
thanks

 

[andrewchang]

yes, it is.

 

[DB]

thanks for all the help andrew