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???

Question

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

Caleb Adams · Accepted Answer

Based on the information I got from you: 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%)#.

Owen Ambrose · Answer

undefined The percent error in the standard enthalpy of reaction for the reaction of zinc with copper sulfate may be large due to inaccuracies in the measurements, experimental errors, or assumptions made during the calculation of the standard enthalpy of reaction. Additionally, variations in experimental conditions or deviations from ideal behavior in the reaction may contribute to the discrepancy between the calculated and expected values.