Two trials were performed. In a calorimeter were combined 10.0 mL of 6.0 M HNO3

Question

Two trials were performed. In a calorimeter were combined 10.0 mL of 6.0 M HNO3 at 25.0°C and 10.0 mL of 1.5 M NaOH at 25.0°C in the first trial. In a second trial, 10.0 mL of 6.0 M HNO3 at 25.0°C and 10 mL of 1.5 M NH3 at 25.0°C were combined in the same calorimeter used in the first trial. Assume that the density of the resultant solution in each trial was 1.05 g/mL, the specific heat of the resultant solution in each trial was 4.20 J/g·°C, and the heat capacity of the calorimeter was 20.0 J/°C. Needed standard enthalpy of formation values can be found in Appendix 4 of the textbook.

a.) Predict the balanced molecular chemical equation for each of the following combinations of reactants:

(1) NaOH(aq) + HNO3(aq)

(2) NH3(aq) + HNO3(aq)

b.) After combining the two solutions, how much would the solution temperature be expected to change in the first trial?

c.) Would you expect the temperature change in Trial 1 to be greater than, less than, equal to, or cannot be determined compared with the temperature change in Trial 2? Justify your choice.

Explanation / Answer

a)

NaOH(aq) + HNO3(aq) = H2O(l) + NaNO3(aq)

NH3(aq) + HNO3(aq) = NH4+(aq) + NO3-(aq)

b)

trial 1:

Q = m*C*(Tf-Ti)

Vtotal = 10+10 = 20 mL

mass = D*V = 20*1.05 = 21 g

HRxn = 21*4.20*(Tf-25)

Qrxn + Qcal = 0

21*4.20*(Tf-25) - 20(Tf-25) =

for HNO3 with NAOH = -55.84 kJ (exothermic)

moles reacted = MV = 1.5*10/1000 = 0.015

Qfrom reaction = n*HXN = 0.015*55.84 = 0.8376 kJ = 837.6 J

21*4.20*(Tf-25) - 20(Tf-25) = 837.6

(21*4.20-20)(Tf-25) = 837.6

Tf = 837.6 /(21*4.20-20) + 25 = 37.28°C

c)

expect this to be slightly lower, since NH3 is a weak base, in comparison with NaOH

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