# 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???

## I don't know what I'm doing... The actual value is $\text{50525.8 cal}$... but I get a huge percent error...

Apr 16, 2016

Based on the information I got from you:

• ${m}_{\text{Zn" = "6.530 g}}$
• m_("CuSO"_4cdot5"H"_2"O") = "5.00 g" (in lab you tend to be handed the hydrate)

From the molar masses of $\text{65.380 g/mol}$ and $\text{249.685 g/mol}$, respectively, we have the following number of $\text{mol}$s:

$6.530 \cancel{\text{g Zn" xx "1 mol Zn"/(65.380 cancel"g Zn") = "0.0999 mol Zn}}$

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 $\text{CuSO"_4cdot5"H"_2"O}$ is the limiting reactant.

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

$\setminus m a t h b f \left(\Delta \text{H"_"rxn" = (q_"rxn")/"mols limiting reactant}\right)$ (1)

Since you performed your reaction at approximately ${25}^{\circ} \text{C}$, this gives you $\Delta {\text{H"_"rxn}}^{\circ}$ as well, given that $\Delta {\text{H" = Delta"H}}^{\circ}$ at ${25}^{\circ} \text{C}$. Given ${q}_{\text{rxn" = "979.72 cal}}$, ${q}_{\text{rxn}}$ is equal to $\text{4099.15 J}$.

Next, let's compare results using $\text{kJ/mol}$, as that is directly solvable using thermodynamic tables as follows:

$\setminus m a t h b f \left(\Delta {\text{H"_"rxn"^@ = sum_R nu_R Delta"H"_(f,R)^@ - sum_P nu_P Delta"H}}_{f , P}^{\circ}\right)$ (2)

My textbook clearly states the enthalpies of formation for ${\text{CuSO}}_{4} \left(s\right)$ and ${\text{ZnSO}}_{4} \left(s\right)$ to be $- 771.4$ and $- \text{982.8 kJ/mol}$, respectively, NOT $\text{cal/mol}$. 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))^@]

$= \left[0 - \text{771.4 kJ/mol"] - [0 - "982.8 kJ/mol}\right]$

= color(blue)("211.4 kJ/mol") ne "211.4 cal". This is why units are very important... That is the value based on $\text{1 mol}$ of ${\text{CuSO}}_{4} \left(s\right)$ in standard conditions (${25}^{\circ} \text{C}$ and $\text{1 bar}$).

When I determine your experimental $\Delta {\text{H"_"rxn}}^{\circ}$ (1), I get:

$\left(4099.15 \cancel{\text{J" xx "1 kJ"/(1000 cancel"J"))/("0.02003 mol CuSO"_4cdot5"H"_2"O}}\right)$

$= \textcolor{b l u e}{\Delta \text{H"_"rxn"^@ = "204.70 kJ/mol}}$

And that is reasonably close to the standard enthalpy of reaction. This is the equivalent of $\text{48924.30 cal/mol}$.

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

From this, using $\text{1 mol}$, I get a percent error of

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

= color(blue)(3.17%).