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 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

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

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