Thermodynamics question homework help?
50.00 mL of 0.20 M lead (II) nitrate solution was added to 30 mL of 1.25 M potassium iodide. These compounds react to form a lead (II) iodide, a bright yellow solid. Both solutions were at room temperature at the start of the reaction (19.6 ˚C). The final temperature of the reaction mixture was 22.2 ˚C.
- What is the theoretical yield of lead (II) iodide?
- What was the q of the solution?
- What was the q of the reaction?
- What was the enthalpy of this reaction?
- What was the mass of water used assuming the density of water was 0.99 g/mL?
50.00 mL of 0.20 M lead (II) nitrate solution was added to 30 mL of 1.25 M potassium iodide. These compounds react to form a lead (II) iodide, a bright yellow solid. Both solutions were at room temperature at the start of the reaction (19.6 ˚C). The final temperature of the reaction mixture was 22.2 ˚C.
- What is the theoretical yield of lead (II) iodide?
- What was the q of the solution?
- What was the q of the reaction?
- What was the enthalpy of this reaction?
- What was the mass of water used assuming the density of water was 0.99 g/mL?
1 Answer
First off, let's write the reaction...
#"Pb"("NO"_3)_2(aq) + 2"KI"(aq) -> "PbI"_2(s) + 2"KNO"_3(aq)#
If
#"50.00 mL"# of#"0.20 M"# #"Pb"("NO"_3)_2# reacts with#"30.00 mL"# of#"1.25 M"# #"KI"# , then we just need the limiting reactant to find the theoretical yield.
#"0.20 mol"/cancel"L" xx cancel"1 L"/(1000 cancel"mL") xx 50.00 cancel"mL" = "0.0100 mols Pb"("NO"_3)_2#
#"1.25 mol"/cancel"L" xx cancel"1 L"/(1000 cancel"mL") xx 30.00 cancel"mL" = "0.0375 mols KI"# Since the mols of
#"KI"# are greater than twice the mols of#"Pb"("NO"_3)_2# , and twice the mols are needed for exact reaction,#"KI"# is in excess and thus#"Pb"("NO"_3)_2# is the limiting reactant.The yield of
#"PbI"_2(s)# is exactly the mols of#"Pb"("NO"_3)_2# from the chemical reaction coefficients, so the theoretical mass yield is:
#color(blue)(m_(PbI_2(s))) = 0.0100 cancel("mols PbI"_2) xx ("461.01 g")/cancel("1 mol PbI"_2) = color(blue)("4.61 g PbI"_2(s))#
The solutions both started at
#19.6^@ "C"# , and absorbed heat from the reaction after mixing to become#22.2^@ "C"# . Hence,#q_(sol n) > 0# .We assume the specific heat capacity of the solution is close enough to that of water,
#"4.184 J/g"^@ "C"# at#25^@ "C"# .Strangely enough, we don't get the density of water until the last question, but we need it here, to approximate the mass of solution formed AFTER mixing.
The heat transferred into the solution is based on the heat capacity (in
#"J/"^@"C"# ) multiplied by the change in temperature (in#""^@ "C"# ), so:
#color(blue)(q_(sol n)) = mC_PDeltaT#
#= (50.00 cancel"mL" + 30.00 cancel"mL soln") xx "0.99 g"/cancel"mL" xx "4.184 J/"cancel"g"cancel(""^@ "C")(22.2cancel(""^@ "C") - 19.6cancel(""^@ "C"))#
#=# #color(blue)("861.6 J")#
By conservation of energy, the heat flow into the solution came out from the reaction, and thus
#color(blue)(q_(rxn) = -q_(sol n) = -"861.6 J")# .
The enthalpy of reaction
#DeltaH_(rxn)# is equal to#q_(rxn)# at constant atmospheric pressure (which we are at on a DAILY basis).I assume it is asked to be in
#"kJ/mol"# (implicitly), and that means it would be of the limiting reactant.
#color(blue)(DeltaH_(rxn)) = -(861.6 cancel"J" xx ("1 kJ")/(1000 cancel"J"))/("0.0100 mols PbI"_2(s))#
#= color(blue)(-"86.16 kJ/mol")#
This feels out of order... we already found it to be a total of
#"80.00 mL"# , and if the density assumed is#"0.99 g/mL"# , this is#color(blue)("79.2 g")# of aqueous solution that we used in#(2)# .
