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