Thursday
July 24, 2014

Homework Help: Chemistry

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 N2(g) + 2H2(g) + Cl2(g) → NH4Cl(s).

We were given the following information:
1)ΔHf for NH3(g) is -46.15 kJ/mol
Formation reaction I came up with: N2(g) + 3/2 H2(g) → NH3(g)

2)ΔHf for HCl(g) is -92.21 kJ/mol
Formation reaction I came up with: H2(g) + Cl2(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 21C, final is 26C (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 21C, final is 16C.

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/gC)(26C-21C)
= 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/gC)(16C- 21C)
= -1046 J

ΔH = -Q/n
= +1.046 kJ/0.075 mol
= +13.95 kJ/mol
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Target equation: N2(g) + 2H2(g) + Cl2(g) → NH4Cl(s)

1)N2(g) + 3/2 H2(g) → NH3(g) ΔHf = -46.15 kJ/mol
2)H2(g) + Cl2(g) → HCl(g) ΔHf = -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


N2(g) + 32 H2(g) → NH3(g)
H2(g) + Cl2(g) → HCl(g)
NH3(g) → NH3(aq)
HCl(g) → HCl(aq)
NH3(aq) + HCl(aq) → NH4Cl(aq)
NH4Cl(aq) → NH4Cl(s)
------------------------------------
N2(g) + 2H2(g) + Cl2(g) → NH4Cl(s)

ΔHf = -46.15 kJ/mol
ΔHf = -92.21 kJ/mol
ΔH = -30.47 kJ/mol
ΔH = -74.78 kJ/mol
ΔH = -41.84 kJ/mol
ΔH = -13.95 kJ/mol
--------------------
ΔHf = -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!!

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