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Topic: Hess' Law  (Read 6521 times)

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Offline Woopy

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Hess' Law
« on: April 12, 2008, 07:48:08 PM »
hello, I am in desperate need of help with this lab my class did!! As you will soon find out I am not a chemistry genius (btw normal chemistry in HS not AP) and we did a lab with these 3 equations, so i'll get this started as quick as possible.

1. NaOH(s) --> NaOH(aq)

2. NaOH(s)  + HCL(aq) --> H2O + NaCL(aq)

3. NaOH (aq) + HCL (aq) --> H2O + NaCL(aq)


now, the question that i'm currently stuck on is this:
Using the 3 equations and your answers to #3, write the thermochemical equations for each part. Use ΔH notation.

well my answers to #3 are

1.ΔH = -5.0 kJ/.103 mol NaOH = ~49 kJ/mole
2.ΔH = -10.0 kJ/.106 mol NaOH = ~94 kJ/mole
3.ΔH = -6.7 kJ/.100 mol NaOH = ~67 kJ/mole
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Offline english

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Re: Hess' Law
« Reply #1 on: April 12, 2008, 07:50:59 PM »
Could you post the question from 3?  There's not much direction from your information you gave.
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Offline Woopy

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Re: Hess' Law
« Reply #2 on: April 12, 2008, 07:55:02 PM »
3. Calculate the ΔH for each part. Leave your answer in kJ/mole
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Offline Woopy

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Re: Hess' Law
« Reply #3 on: April 12, 2008, 07:59:09 PM »
i got question 4 (onyl cuz i asked on WoW forums and got a reply)

now #5 saids

Using Hess' Law, which 2 thermochemical equations add up to the other? DO NOT flip equations! Prove it by using equations and be sure to label each equation. use ΔH notation
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Offline Woopy

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Re: Hess' Law
« Reply #4 on: April 12, 2008, 08:07:54 PM »
come back g english, i need your brain!
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Offline Woopy

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Re: Hess' Law
« Reply #5 on: April 12, 2008, 08:38:23 PM »
can anyone help? I am so dang dumb in this field and I just don't know what to do
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Offline english

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Re: Hess' Law
« Reply #6 on: April 12, 2008, 08:53:22 PM »
This is very simple.  You're over-thinking it.  What's the difference between a thermochemical equation and a chemical equation?
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Offline Woopy

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Re: Hess' Law
« Reply #7 on: April 12, 2008, 09:01:50 PM »
one has to do with temperature, but please understand that your working with me, and I am still on the first question after almost 2 hours.

NaOH (s) + 49 kj/mol ---> NaOH (aq)
-425.9 kj/mol + 49 kj/mol 
ΔH= -376.9 kJ/mol

NaOH (s) + HCL (aq) + 94 kJ/mol ---> H2O + NaCL (aq)
-425.9 kj/mol + X + 94 kJ/mol ---> -285.8 kj/mol + Y
ΔH = ???

NaOH (aq) + HCL (aq) +67 kJ/mol ---> H2O + NaCL (aq)
X + Y + 67 kJ/mol ---> -285.8 kJ/mol + Z
ΔH = ???

the heat of formation chart doesnt give the value for the stuff i used variables for
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Offline Woopy

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Re: Hess' Law
« Reply #8 on: April 12, 2008, 09:13:15 PM »
g english am I doing this right or is it wrong?
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Offline Woopy

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Re: Hess' Law
« Reply #9 on: April 12, 2008, 09:37:46 PM »
plz someone help, i've attempted the problem and im still in this struggle!
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Offline Woopy

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Re: Hess' Law
« Reply #10 on: April 12, 2008, 10:30:57 PM »
after 3.5 hours, I still cant do this. is there no heat formation of HCl (aq) or NaOH (aq) ?? please, someone out there help
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Offline AWK

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Re: Hess' Law
« Reply #11 on: April 14, 2008, 01:40:48 AM »
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