Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Mass of beaker plus solid Mass of water, m Mass of solid, m Initial temperature

ID: 570905 • Letter: M

Question

Mass of beaker plus solid Mass of water, m Mass of solid, m Initial temperature Final temperature tin,o for the reaction (Eq 1) (S.H-4.18 /g'C) AHI for the reaction (Eq,5) The quantity you have just calculated is approximately equal to the heat of solution of your sample.Calculate 11.295 . 999 53.tn .11 13 ..4 89 joules the heat of solution per gram of solid sample. r92 4.qq9 The solution reaction is (Underline the correct answer.) Give your reasoning. amp Sold unknown hDs Otni Formula of compound used (if furnished)SMolar mass Heat of solution per mole of compound C. Heat of Neutralization gmol Initial temperature of HCI solution Initial temperature of NaOH solution Final temperature of neutralized mixture Change in temperature , (use the average ofthe initial temperatures of 3/ "C HCI and NaOH) ho (assume 50.0 mL of solution and use average density of 1.02 g/mL.) Total AH for the neutralization reaction kJ/mol H per mole of H' and OH ions reacting "The value of &H; will be approximaie for several reasons One is that we do aot include the amoons of hea ahsorbed by the soae. This effect is smaller than the Skely experimental emx, and thas we wil ignore it

Explanation / Answer

Part A)

Total mass = 48.298 + 4.999 = 53.297 g

Specific heat of water = 4.18 J/g.oC

dT = (final - initial)temperature

     = 13 - 20 = -7oC

So,

q(H2O) = mCpdT

             = 53.297 x 4.18 x -7 = -1559.47 J

dHrxn = 1559.47 J

dH(soln) = 1559.47 J

Identity of solid = KNO3

molar mass KNO3 = 101.1 g/mol

moles of KNO3 added = 4.999 g/101.1 g/mol = 0.04945 mol

dH(soln) = 1559.47 J/0.04945 mol x 1000 = 31.54 kJ/mol

C. Heat of neutralization

density = 1.02 g/ml

mass of solution = 50 ml x 1.02 g/ml = 51 g

Cp = 4.18 J/g.oC

average initial temperature = (20 + 19)/2 = 19.5 oC

Final temperature = 31 oC

dT = (final - initial) temperature

      = 31 - 19.5 = 11.5 oC

q(H2O) = mCpdT

             = 51 x 4.18 x 11.5 = 2451.57 J

dH for the neutralization reaction = -2451.57 J

dH(kJ/mol) = -2451.57/moles of H+(or OH-) in solution x 1000