Quiz 1 Prep #5

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Diwana Lucero 3K
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Joined: Fri Jul 15, 2016 3:00 am

Quiz 1 Prep #5

Postby Diwana Lucero 3K » Wed Jan 25, 2017 5:20 pm

Calculate the standard reaction enthalpy for the reaction: N2H4(ℓ) + H2(g) → NH3(g)
Given:
N2H4(ℓ) + O2(g) → N2(g) + 2H2O(g) ∆H◦ = −543 kJ · mol −1
2 H2(g) + O2(g) → 2 H2O(g) ∆H◦ = −484 kJ · mol −1
N2(g) + 3 H2(g) → 2 NH3(g) ∆H◦ = −92.2 kJ · mol −1

I worked out this problem but I got N2H4(ℓ) + H2(g) → 2 NH3(g). I'm not sure how to get rid of the coefficient 2 in NH3, unless this is a typo?

Mimi Lec 2H
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Joined: Fri Jul 22, 2016 3:00 am

Re: Quiz 1 Prep #5

Postby Mimi Lec 2H » Wed Jan 25, 2017 5:54 pm

I'm almost positive it's a typo and there has to be a coefficient of 2 for NH3. Without it, the reaction we are trying to find the standard enthalpy for is unbalanced.

N2H4(ℓ) + H2(g) → NH3(g) : the reactant side has 2 nitrogen, 6 hydrogen and the product side has 1 nitrogen, 3 hydrogen, so there should be a coefficient of 2

Alex_Wagas_3K
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Joined: Tue Nov 25, 2014 3:00 am

Re: Quiz 1 Prep #5

Postby Alex_Wagas_3K » Wed Jan 25, 2017 6:34 pm

As mentioned above, it is probably a typo as without the coefficient of 2 NH3, the chemical equation itself would be unbalanced.

Melody Li 2L
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Joined: Fri Jul 22, 2016 3:00 am

Re: Quiz 1 Prep #5

Postby Melody Li 2L » Fri Jan 27, 2017 8:30 pm

^^ Like said above, not sure if there was an error in the equation but it was not originally balanced therefore the original reaction should be: N2H4(l) + H2(g) -> 2NH3(g) once you balance it .From here, your equation has 2NH3 in the product side and your end reaction should match the original reaction.

Nasir_Ahmed
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Joined: Wed Sep 21, 2016 2:58 pm

Re: Quiz 1 Prep #5

Postby Nasir_Ahmed » Sat Jan 28, 2017 2:28 pm

There's no error. You always have to balance the equation before you even get started. So a final product of 2NH3 is correct.

Diwana Lucero 3K
Posts: 26
Joined: Fri Jul 15, 2016 3:00 am

Re: Quiz 1 Prep #5

Postby Diwana Lucero 3K » Sun Jan 29, 2017 4:02 pm

Okay, thank you guys!


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