## 4E.5, 4E.7

Reina Robles 2B
Posts: 71
Joined: Fri Aug 09, 2019 12:16 am

### 4E.5, 4E.7

I'm getting almost all of the answers wrong for these two questions and I'm not sure why. Any help is appreciated, thanks!

4E.5
a) 3C2H2 (g) --> C6H6 (g)
What I did:
[3(837)] + [6(-348)] = 423 kJ.mol-1

4E.7
a) N2 (g) + 3F2 (g) --> 2NF3 (g)
[944+(3*158)] + [6(-270)] = -169 kJ.mol-1

b) CH3CHCH2 (g) + H20 (g) --> CH3CH(OH)CH3 (g)
[612+463] + [-348-412] = 315 kJ.mol-1

c) CH4 (g) + Cl2 (g) --> CH3Cl (g) + HCl (g)
[412+242] + [-338-421] = 273 kJ.mol-1

AArmellini_1I
Posts: 107
Joined: Fri Aug 09, 2019 12:15 am

### Re: 4E.5, 4E.7

4E.5
3(837)kJ/mol - 6(518)kJ/mol = -597 kJ/mol
4E.7
a)
1(944)kJ/mol + 3(158)kJ/mol - 6(270)kJ/mol = -202kJ/mol
b)
1(463)kJ/mol + 1(612)kJ/mol - 1(412)kJ/mol - 1(360)kJ/mol - 1(348)kJ/mol = -45kJ/mol
c)
1(412)kJ/mol + 1(242)kJ/mol - 1(338)kJ/mol - 1(431)kJ/mol = -115 kJ/mol

Hope this helps!

Luc Zelissen 1K
Posts: 57
Joined: Mon Jun 17, 2019 7:23 am

### Re: 4E.5, 4E.7

For 4.E.5,
Looking at the lewis structure for C2H2 gives us 1 C-C triple bond, which has an enthalpy of 837, and also 2 C-H bonds which have a combined value of 824. Add these 2 numbers and multiply by the moles, which is 3, and you get the total enthalpy for C2H2, 4983. For benzene, the table of bond enthalpies makes a special note about benzene and its C-C bond resonance structures which gives an altered value of 518, which is then multiplied by 6 because of the 6 bonds. It also has 6 C-H bonds, each of which has 412, which gives the total enthalpy of benzene of -5580. Then you add these 2 values up and get 4983+(-5580)=-597

Justin Vayakone 1C
Posts: 110
Joined: Sat Sep 07, 2019 12:19 am

### Re: 4E.5, 4E.7

For 4E.5) you used the C-C bond enthalpy (-348kJ/mol) when it should be the C-H bond enthalpy which is 518kJ/mol

For 4E.7)
a) Your work looks right. You should be getting -202 kJ/mol. Maybe you plugged it into your calculator wrong?
b) You forgot to include the bond enthalpy of C-C in the products. Since you included the C=C double bond in the reactants, a C-C bond has to be formed to reconnect the two carbons.
c) Your work "[412+242] + [-338-421]" should equal -105 kJ/mol, but you say you got 273 kJ/mol. Plugged it into the calculator wrong again? Anyways, the only mistake here in your work is that the bond enthalpy of H-Cl bond is 431 kJ/mol, not 421. So the answer should be [412+242] + [-338-431] = -115 kJ/mol.

Reina Robles 2B
Posts: 71
Joined: Fri Aug 09, 2019 12:16 am

### Re: 4E.5, 4E.7

Thank you!

Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

### Who is online

Users browsing this forum: No registered users and 2 guests