# 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???

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I don't know what I'm doing... The actual value is #"50525.8 cal"# ... but I get a huge percent error...

I don't know what I'm doing... The actual value is

##### 1 Answer

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

#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 **limiting reactant**.

Now, your stated value of **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

Next, let's compare results using

#\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 **(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 **(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

On the other hand, the standard enthalpy of reaction I got is equal to

From this, using **percent error** of

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

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