8.67

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Rakhi Ratanjee 1D
Posts: 59
Joined: Fri Sep 29, 2017 7:04 am

8.67

Postby Rakhi Ratanjee 1D » Sun Jan 21, 2018 9:14 pm

For the reaction 1H2+(1/2)O2--> 1H2O(liquid), the final calculation of the bond enthalpies comes to 242kJ/mol. From the tables in the textbook, the answer would work out to be a +242kJ/mol, but the solutions manual states that it is negative. Why?
Also, why do you have subtract 44 kJ/mol to get the final answer?

sahajgill
Posts: 20
Joined: Fri Sep 29, 2017 7:06 am

Re: 8.67

Postby sahajgill » Sun Jan 21, 2018 9:21 pm

I did this question by using the bond enthalpies. The total enthalpy required to break the bonds was 684 kj.mol^-1 but the total enthalpy of the formation of the bonds was -926 kj.mol^-1 and is negative since when bonds form, it is an exothermic process. Add -926 to 684 and you get -242 kj.mol^-1

Rakhi Ratanjee 1D
Posts: 59
Joined: Fri Sep 29, 2017 7:04 am

Re: 8.67

Postby Rakhi Ratanjee 1D » Sun Jan 21, 2018 9:30 pm

In 8.67 part B, the desired reaction is C(gr)+2H2(g)+(1/2)O2(g) --> CH3OH(l)
What is C(gr)?

Rakhi Ratanjee 1D
Posts: 59
Joined: Fri Sep 29, 2017 7:04 am

Re: 8.67

Postby Rakhi Ratanjee 1D » Tue Jan 23, 2018 1:00 pm

C(gr) just means that the natural form of carbon is in solid graphite.


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