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April 19, 2014

April 19, 2014

Posted by **a Canadian** on Wednesday, February 20, 2013 at 6:27pm.

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 7:30pmOh, 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 9:34pmI didn't check any of the steps you went through. I just scrolled down and answered the percent part.

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