I measured `"5.00 g"` copper sulfate solid and `"6.530 g"` zinc. If for this reaction `q_(rxn) = "979.72 cal"` of energy was involved, why is my percent error of the standard enthalpy of reaction for the reaction of zinc with copper sulfate so big???

Based on the information I got from you:

• `m_"Zn" = "6.530 g"`
• `m_("CuSO"_4cdot5"H"_2"O") = "5.00 g"` (in lab you tend to be handed the hydrate)

From the molar masses of `"65.380 g/mol"` and `"249.685 g/mol"`, respectively, we have the following number of `"mol"`s:

`6.530 cancel"g Zn" xx "1 mol Zn"/(65.380 cancel"g Zn") = "0.0999 mol Zn"`

`5.00 cancel("g CuSO"_4cdot5"H"_2"O") xx ("1 mol CuSO"_4)/(249.685 cancel("g CuSO"_4cdot5"H"_2"O")) = "0.02003 mol CuSO"_4cdot5"H"_2"O"`

So, we know now that `"CuSO"_4cdot5"H"_2"O"` is the limiting reactant.

Now, your stated value of `"979.72 cal"` that you claim is correct, which you have stated before as `q_"rxn"`, is NOT the same as `Delta"H"_"rxn"` unless you are at a constant pressure. If you are at a constant pressure, then:

`\mathbf(Delta"H"_"rxn" = (q_"rxn")/"mols limiting reactant")` (1)

Since you performed your reaction at approximately `25^@ "C"`, this gives you `Delta"H"_"rxn"^@` as well, given that `Delta"H" = Delta"H"^@` at `25^@ "C"`. Given `q_"rxn" = "979.72 cal"`, `q_"rxn"` is equal to `"4099.15 J"`.

Next, let's compare results using `"kJ/mol"`, as that is directly solvable using thermodynamic tables as follows:

`\mathbf(Delta"H"_"rxn"^@ = sum_R nu_R Delta"H"_(f,R)^@ - sum_P nu_P Delta"H"_(f,P)^@)` (2)

My textbook clearly states the enthalpies of formation for `"CuSO"_4(s)` and `"ZnSO"_4(s)` to be `-771.4` and `-"982.8 kJ/mol"`, respectively, NOT `"cal/mol"`. Therefore, we should get (2):

`color(blue)(Delta"H"_"rxn"^@) = [Delta"H"_(f,"Zn"(s))^@ + Delta"H"_(f,"CuSO"_4(s))^@] - [Delta"H"_(f,"Cu"(s))^@ + Delta"H"_(f,"ZnSO"_4(s))^@]`

`= [0 - "771.4 kJ/mol"] - [0 - "982.8 kJ/mol"]`

`= color(blue)("211.4 kJ/mol") ne "211.4 cal"`. This is why units are very important... That is the value based on `"1 mol"` of `"CuSO"_4(s)` in standard conditions (`25^@ "C"` and `"1 bar"`).

When I determine your experimental `Delta"H"_"rxn"^@` (1), I get:

`(4099.15 cancel"J" xx "1 kJ"/(1000 cancel"J"))/("0.02003 mol CuSO"_4cdot5"H"_2"O")`

`= color(blue)(Delta"H"_"rxn"^@ = "204.70 kJ/mol")`

And that is reasonably close to the standard enthalpy of reaction. This is the equivalent of `"48924.30 cal/mol"`.

On the other hand, the standard enthalpy of reaction I got is equal to `"211.4 kJ/mol" xx "1000 J"/"1 kJ" xx "1 cal"/"4.184 J" = "50525.8 cal/mol"` (which is NOT equal to `"50525.8 cal"`!!!).

From this, using `"1 mol"`, I get a percent error of

`(|"48924.30 cal" - "50525.8 cal"|)/("50525.8 cal")xx100%`

`= color(blue)(3.17%)`.