Chemistry

posted by .

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 -

    % 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 -

    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 -

    I didn't check any of the steps you went through. I just scrolled down and answered the percent part.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chemistry questions

    how do you distinquish between a limiting reactant and the excess reactant in a chemical reaction?
  2. Chemistry

    I am trying to find the theoretical yield for this lab I am working on. The steps to finidng it would be great. In my lab I took 40mls of H2O + 4g of AgNO3+ 8g of NaCL then I stirred it and filtered in a buchner funnel to be left with …
  3. Chemistry

    I am trying to find the theoretical yield for this lab I am working on. The steps to finidng it would be great. In my lab I took 40mls of H2O + 4g of AgNO3+ 8g of NaCL then I stirred it and filtered in a buchner funnel to be left with …
  4. Chemistry

    Calculate the theoretical pH after 2.50 mL and 9.50 mL of NaOH has been added in both the titration of HCl and of HC2H3O2. Indicate if the volume of NaOH is before or after the equivalence point. Compare these theoretical values with …
  5. Chemistry

    It's a molar heat reaction lab. 0.07g Mg Mg+2HCl->MgCl2+H2 What is the actual value for the heat of reaction based on the enthalpy's of formation?
  6. Chemistry! Test tomorrow, pls help !

    we're given the equation: C6H6(l)+Br2(l) --> C6H5Br(l)+HBr(g) you need to prepare 50.0g pf bromobenzene and you expect no more than a 75% yield. How much benzene should you begin with if the yield is 75%. i know that % yield = (actual/theoretical) …
  7. Chemistry

    I did a lab today in class and part of the lab is to experimentally find the enthalpy for the dissolution of ammonium chloride by dissolving 4.0 g of NH4Cl(s) in 50.0 mL of H2O(l). I got a -5°C temperature change and calculated delta …
  8. Chemistry

    Why wouldn't you do this reaction directly in a lab (and would rather use Hess' Law to determine the enthalpy of formation)?
  9. English

    This is a chemistry lab report, but I was wondering if someone could edit the grammar and look at the sentence structure, etc. I know it's really long and boring, but even if you just look over a bit of it, I would really appreciate …
  10. Chemistry

    For an enthalpy lab, why would calculated value differ from the theoretical value (in percent error) for the heat reaction of Mg/HCl and MgO/HCl

More Similar Questions