Posted by a Canadian on Wednesday, February 20, 2013 at 6:27pm.
Can you check my calculations, please? It's for a lab we did in order to find the the enthalpy of formation of NH4Cl(s). My final answer was -299.4 kJ, while the theoretical, or actual, value is -314.4 kJ. It was the closest value out of everyone in the entire class. I'm not bragging, I actually think it's a bit too good to be true! I'll be the first to admit that my labs usually do not turn out very well!
I don't know if you necessarily need to know this stuff to check the calculations, but in case, here it is (if not, you can just skip down to where it says "CALCULATIONS").
The formation reaction of NH4Cl(s) is ½N2(g) + 2H2(g) + ½Cl2(g) → NH4Cl(s).
We were given the following information:
1)ΔH°f for NH3(g) is -46.15 kJ/mol
Formation reaction I came up with: ½N2(g) + 3/2 H2(g) → NH3(g)
2)ΔH°f for HCl(g) is -92.21 kJ/mol
Formation reaction I came up with: ½H2(g) + ½Cl2(g) → HCl(g)
3)NH3(g) → NH3(aq) ΔH° = -30.47 kJ/mol
4)HCl(g) → HCl(aq) ΔH° = -74.78 kJ/mol
5)NH3(aq) + HCl(aq) → NH4Cl(aq) ΔH° = ?
6)NH4Cl(s) → NH4Cl(aq) ΔH° = ?
experimentally, we found that the for
5)NH3(aq) + HCl(aq) → NH4Cl(aq), initial temperature is 21°C, final is 26°C (25 mL of 1.0M for both). For 6)NH4Cl(s) → NH4Cl(aq), (4.0g of ammonium chloride, 50 mL of H2O) initial temperature is 21°C, final is 16°C.
CALCULATIONS:
NH3(aq) + HCl(aq) → NH4Cl(aq)
25 mL 25 mL
1.0 mol/L 1.0 mol/L
n = c x v
= 1 mol/L x 0.025L
= 0.025 mol
Q = mc ΔT
=(25g +25g)(4.184 J/g°C)(26°C-21°C)
= 1046 J
ΔH = -Q/n
= -1.046 kJ/0.025 mol
= -41.84 kJ/mol
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
NH4Cl(s) → NH4Cl(aq)
4.0 g
n = m/MM
= 4.0 g/(14.01 + 4.04+35.45)g/mol
= 4.0g/53.5g/mol
= 0.075 mol
Q = mc ΔT
= (50g)(4.184 J/g°C)(16°C- 21°C)
= -1046 J
ΔH = -Q/n
= +1.046 kJ/0.075 mol
= +13.95 kJ/mol
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Target equation: ½N2(g) + 2H2(g) + ½Cl2(g) → NH4Cl(s)
1)½N2(g) + 3/2 H2(g) → NH3(g) ΔH°f = -46.15 kJ/mol
2)½H2(g) + ½Cl2(g) → HCl(g) ΔH°f = -92.21 kJ/mol
3)NH3(g) → NH3(aq) ΔH° = -30.47 kJ/mol
4)HCl(g) → HCl(aq) ΔH° = -74.78 kJ/mol
5)NH3(aq) + HCl(aq) → NH4Cl(aq)ΔH° = -41.84 kJ/mol
6)NH4Cl(s) → NH4Cl(aq) ΔH° = +13.95 kJ/mol
½N2(g) + 32 H2(g) → NH3(g)
½H2(g) + ½Cl2(g) → HCl(g)
NH3(g) → NH3(aq)
HCl(g) → HCl(aq)
NH3(aq) + HCl(aq) → NH4Cl(aq)
NH4Cl(aq) → NH4Cl(s)
------------------------------------
½N2(g) + 2H2(g) + ½Cl2(g) → NH4Cl(s)
ΔH°f = -46.15 kJ/mol
ΔH°f = -92.21 kJ/mol
ΔH° = -30.47 kJ/mol
ΔH° = -74.78 kJ/mol
ΔH° = -41.84 kJ/mol
ΔH° = -13.95 kJ/mol
--------------------
ΔH°f = -299.4 kJ/mol
Percent Error:
% error = |experimental value – theoretical value| / theoretical value
= |-299.4 kJ/mol – (-314.4 kJ/mol)| / -314.4 kJ/mol
Is the %error equation I used correct? Because when I looked at the internet it gave me |ev-tv|/tv as well as |tv-ev|/tv. Are they the same thing?
Thanks so much in advance!!
- Chemistry - DrBob222, Wednesday, February 20, 2013 at 7:21pm
% error = |experimental value – theoretical value| / theoretical value
= |-299.4 kJ/mol – (-314.4 kJ/mol)| / -314.4 kJ/mol
You have calculated the fraction; if you want the percent you must multiply by 100.
Is the %error equation I used correct? Because when I looked at the internet it gave me |ev-tv|/tv as well as |tv-ev|/tv. Are they the same thing?
Placing the |...| around the difference between ev and tv means it's the absolute difference (irrespective of sign) so it makes no difference. Some like to write it ev-tv so that the SIGN shows that the error was positive or negative. That is suppose actual value is 21 and experimental value is 20, then (20-21)/21 gives a negative number meaning the exp value was less than the actual value. If the values were reversed, you would have 21-20/20 and the fraction would mean the exp value was greater than the actual value. I like the absolute value way of doing it; you don't worry about which way, just how much. Your 5% error is pretty good. Congratulations.
- Chemistry - a Canadian, Wednesday, February 20, 2013 at 7:30pm
Oh, oops ! I forgot to put x100 and then the rest of my answer, 4.8% (or 5%)
I think it's a small miracle that I didn't make any mistakes when calculating the enthalpy of formation, there were a ton of steps!
Thank you so much!
- Chemistry - DrBob222, Wednesday, February 20, 2013 at 9:34pm
I didn't check any of the steps you went through. I just scrolled down and answered the percent part.
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