Acoffee cup calorimeter with a heat capacity of 4.70 JrCwas used to measure the
ID: 522303 • Letter: A
Question
Acoffee cup calorimeter with a heat capacity of 4.70 JrCwas used to measure the Ghange in enthalpy of a precipitation reaction. A50.0 mL solution of 0.360 M AgNO3 was mixed with 50.0 mL of 0.200 M KBr. After mixing, the temperature was observed to increase by 1.94 C. Calculate the enthalpy of reaction, AHnen, per mole of precipitate formed gBr). Assume the specific heat of the product solution is 4.18 Jl(g. "C)and that the density of both the reactant solutions is 1.00 glmt. Calculate the theoretical moles of precipitate formed from AgNO3 (left) and KBr (right). Number Number O moles moles Number Tools Calculate the heat ch d by the calorimeter contents, qoontents. L 10 Number Calculate the heat change experienced by the calorimeter, qoal. U Number Calculate the heat change produced by the solution process, qsolution. using the mole values calcuated above, calulate 4Hsolution for one mole of precipitate formed Number kJ moleExplanation / Answer
The balanced chemical equation for the precipitation reaction is
AgNO3 (aq) + KBr (aq) ------> AgBr (s) + KNO3 (aq) …..(1)
As per the balanced stoichiometric reaction,
1 mole AgNO3 = 1mole KBr = 1 mole AgBr
Moles of AgNO3 = (volume of AgNO3 in L)*(concentration of AgNO3 in mol/L) = (50.0 mL)*(1 L/1000 mL)*(0.360 mol/L) = 0.018 mole (1 M = 1 mol/L) (ans).
Moles of KBr = (volume of KBr in L)*(concentration of KBr in mol/L) = (50.0 mL)*(1 L/1000 mL)*(0.200 mol/L) = 0.01 mol (ans).
Volume of the contents in the calorimeter = (50.0 + 50.0) mL = 100.0 mL; density of reactant solutions = 1.00 g/mL.
Therefore, mass of the reactant solutions = (100.0 mL)*(1.00 g/mL) = 100.0 g.
Specific heat of the product solution = 4.18 J/(g.C).
Therefore, heat change produced by the calorimeter contents, qcontents = (mass of the solution)*(specific heat of the solution)*(temperature change) = (100.0 g)*(4.18 J/g.C)*(1.94 C) = 810.92 J (ans).
Heat change experienced by the calorimeter, qcal = (heat capacity of the calorimeter)*(change in temperature) = (4.70 J/C)*(1.94 C) = 9.118 J (ans).
The heat change produced by the solution process is qsolution = qcontents + qcal = (810.92 J) + (9.118 J) = 820.038 J (ans).
The solution provides the enthalpy required for the precipitation reaction; therefore, the enthalpy of the solution is Hsolution = -qsolution = -820.038 J (ans).
We note that AgNO3 is present in excess (refer to the equation and the discussion above). KBr is the limiting reactant and the mass of the product formed is decided by the mass of KBr.
Moles of AgBr formed = (0.01 mole KBr)*(1 mole AgBr/1 mole KBr) = 0.01 mole AgBr.
Therefore, the enthalpy of the reaction is
Hrxn = Hsolution/(moles of precipitate formed) = (-820.038 J)/(0.01 mole) = 82003.8 J/mol = (82003.8 J/mol)*(1 kJ/1000 J) = -82.0038 kJ/mol -82.0 kJ/mol (ans).
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