thermochemistry question

[wisdom_tooth]

Calculate the enthalpy change for the following reaction:

 

C(s) + 2H2(g) ----> CH4 (g)

 

(Bond dissociation energies of H-H and C-H bonds are 436 KJ/mol and 415 KJ/mol, respectively. Delta H standard of formation of C(g)= 715 KJ/mol)

 

 

My solution:

 

Delta H of rxn= (Delta H of bonds broken) + (Delta H of bonds formed)

equivalent to

Delta H of rxn= (Total energy input) - (Total energy released)

 

so two H-H bonds were broken therefore +2(415)

C(s) ----> C(g) (in methane gas) therefore that's just +715

and finally 4 C-H bonds were formed to make methane -4(436)

 

therefore...

 

Delta H of rxn=[2(415)+715]-[4(436)]

 

is this correct??

 

I'm pretty confident this is how this question is solved yet for some reason in my kaplan Lesson book (seeing that the teacher went so fast lol) i had something like this

 

Delta H of rxn=[2(436)+715]-[4(436]

I'm pretty sure it's wrong, maybe during the rush of things instead of writing down 2(415) I wrote down 2(436).

 

If anyone can check if I did this right it would be greatly appreciated

 

thanks,

-Wisdom_tooth

[wisdom_tooth]

btw I'd like to point out that I really really really hate thermochemistry LOL

 

[Kuantum]

Hi wisdom,

 

I'm looking at your solution and it appears that you switched the values for the bond enthalpies of the H-H and C-H bond. Also, in the Kaplan solution, it appears that they used the same value of 436 for the H-H and C-H bond enthalpies.

 

I think the answer is as follows

 

= 2(436) + 715 - 4(415)

 

I hope that helps.

[avenir001]

...except I think the enthalpy change is calculated by subtracting bond dissociation energies of the reactants from those of the products...so if you want to get the sign of the enthalpy change right, it would be more like

4(415)-2(436)-715 which comes out positive, meaning you have to put energy into the system to go from left to right.

hmm...anyone disagree?

[wisdom_tooth]

Hi wisdom,

 

I'm looking at your solution and it appears that you switched the values for the bond enthalpies of the H-H and C-H bond. Also, in the Kaplan solution, it appears that they used the same value of 436 for the H-H and C-H bond enthalpies.

 

I think the answer is as follows

 

= 2(436) + 715 - 4(415)

 

I hope that helps.

so is that a mistake on kaplan's part?

[Kuantum]

both you and Kaplan according to what you posted

[wisdom_tooth]

...except I think the enthalpy change is calculated by subtracting bond dissociation energies of the reactants from those of the products...so if you want to get the sign of the enthalpy change right, it would be more like

4(415)-2(436)-715 which comes out positive, meaning you have to put energy into the system to go from left to right.

hmm...anyone disagree?

I'm actually a little bit confused lol, so don't quote me on what about to say but this is how I see it: you're right about the fact that delta H= delta H of the products - delta H of the reactants but in the case of bond dissociation it's different.

Delta H of the reaction= (Delta H of bonds broken)+(Delta H of bonds formed)

we know that to break a bond, you put in energy and to form a bond to release energy. So therefore Delta H of the rxn= Total energy input (so the energy required to break bonds) - Total energy release(the amount of energy to form bonds). If you took the Kaplan course and have the kaplan lesson book, this info is found on page 274. In the problem that I have mentioned, I see it as follows, H2 (2 moles of it) are broken therefore the value will be positive, as well the formation of solid C to gaseous C is also a postive value (just use the standard heat of formation value of 715) finally, the product we have is methane and methane is made up of 4 C-H bonds and we know that to make a bond you actually release energy hence why its value will be negative. Okay that was a mouthful, i hope that makes sense!

[wisdom_tooth]

both you and Kaplan according to what you posted

ohhh lol I get it, i switched 436 and 415

I understand now lol

I feel like an idiot

 

It's not so much kaplan's mistake, I think I made a mistake writing down the solution because my teacher said it verbally (really fast)

thanks for your help

[avenir001]

I'm actually a little bit confused lol, so don't quote me on what about to say but this is how I see it: you're right about the fact that delta H= delta H of the products - delta H of the reactants but in the case of bond dissociation it's different.

Delta H of the reaction= (Delta H of bonds broken)+(Delta H of bonds formed)

we know that to break a bond, you put in energy and to form a bond to release energy. So therefore Delta H of the rxn= Total energy input (so the energy required to break bonds) - Total energy release(the amount of energy to form bonds). If you took the Kaplan course and have the kaplan lesson book, this info is found on page 274. In the problem that I have mentioned, I see it as follows, H2 (2 moles of it) are broken therefore the value will be positive, as well the formation of solid C to gaseous C is also a postive value (just use the standard heat of formation value of 715) finally, the product we have is methane and methane is made up of 4 C-H bonds and we know that to make a bond you actually release energy hence why its value will be negative. Okay that was a mouthful, i hope that makes sense!

 

Oh right!..that makes sense...thanks.

Of course I knew it...just wanted you to reason through it & learn by explaining it to me;)

Kidding.