8.63

Moderators: Chem_Mod, Chem_Admin

Josh Moy 1H
Posts: 58
Joined: Thu Jul 13, 2017 3:00 am

8.63

Postby Josh Moy 1H » Wed Jan 24, 2018 10:32 am

For part c does the h20 (1) have a different formation enthalpy because when i calculate evreything my value is off

Lily Sperling 1E
Posts: 49
Joined: Tue Oct 10, 2017 7:14 am

Re: 8.63

Postby Lily Sperling 1E » Wed Jan 24, 2018 10:43 am

The enthalpy of formation for H2O(l) is -285.83, but you would have to multiply that number by 2 because there are 2 moles in the reaction.

Jingyi Li 2C
Posts: 56
Joined: Fri Sep 29, 2017 7:06 am

Re: 8.63

Postby Jingyi Li 2C » Wed Jan 24, 2018 10:44 am

The standard enthalpy of formation for H2O(l) should be -285.83 kJ/mol.

standard reaction enthalpy = [(-471.5 kJ/mol) * 1 mol + 2 mol * (-285.83 kJ/mol)] - [(-39.7 kJ/mol) * 1 mol + 2 mol * (-482.37 kJ/mol)]
=-38.72 kJ

manasa933
Posts: 72
Joined: Fri Sep 29, 2017 7:04 am

Re: 8.63

Postby manasa933 » Wed Jan 24, 2018 10:45 am

The enthalpy of formation of liquid in it's liquid state is -285.85 kJ/mol

Cassandra Mullen 1E
Posts: 54
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 2 times

Re: 8.63

Postby Cassandra Mullen 1E » Fri Jan 26, 2018 7:58 am

There is a mistake in the solution manual as Dr. Lavelle pointed out on his website: the enthalpy of formation of K2S is actually -471.5 kJ/mol but it says -417.5 kJ/mol in the manual. The correct answer is then -38.72 kJ.


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 1 guest