text problem 4J.5

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JosephineF
Posts: 49
Joined: Wed Sep 18, 2019 12:17 am

text problem 4J.5

Postby JosephineF » Mon Mar 02, 2020 1:49 pm

it is asking to write the balanced formation reaction for NH3. The solution manual says 1/2N2 + 3/2H2 --> NH3. Is there a formula for how to arrive at these numbers or is it just guess and check? I'm confused as to how 2 Ns goes to only 1 N and how H2 gains another H

Parker Smith
Posts: 102
Joined: Thu Jul 25, 2019 12:15 am

Re: text problem 4J.5

Postby Parker Smith » Mon Mar 02, 2020 2:11 pm

the (1/2) N2 revers to only using a half mole of N2 (literally only one nitrogen molecule) to make one mole of NH3, which makes sense. Same applies with using (3/2) of the diatomic hydrogen to make one mole of NH3 (since 2x3=6 and 6/3=3, which lines up with the 3 hydrogen molecules in one mole of NH3)

Ryan Yee 1J
Posts: 101
Joined: Sat Aug 17, 2019 12:16 am

Re: text problem 4J.5

Postby Ryan Yee 1J » Wed Mar 11, 2020 8:23 pm

If you count up the total molecules in these equations, both sides have equal numbers of the required atoms. The 1/2(N2) means that there is only 1 nitrogen atom, and the 3/2(H2) means there are 3 hydrogen atoms, so they both come together to make 1(NH3)


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