3 (20 PTS). (a) From the data, plot the enthalpy of 1 mole of solution at 27°C a

Question

3 (20 PTS). (a) From the data, plot the enthalpy of 1 mole of solution at 27°C as a function of the weight percent HNOs. Use as reference states liquid water at 0°C and liquid HNO3 at 0°C. You can assume that Cp for H2O is 75 J/g mol .°C and for HNO3, 125 J/g mol.°C 8 -AHsoln at 27°C Moles H20 added (J/g mol HNO3) To 1 mole HNO3 3350 5440 6900 8370 10880 14230 17150 20290 24060 25940 27820 30540 31170 0.0 0.1 0.2 0.3 0.5 0.67 1.0 1.5 2.0 3.0 4.0 5.0 10.0 20.0 b) Compute the energy absorbed or evolved at 27 °C on making a solution of 4 moles of HNOs, and 4 moles of water by mixing a solution of 33 1/3 mol % acid with one of 60 mol % HNO3. 8 8

Explanation / Answer

64.8

Weight of HNO3 is taken to be 63 grams per mole

And for H2O weight is 18 grams per mole

Positive signs indicates heat is released and Negative signs are for Heat absorbed

Heat required to raise the temperature of solution from 0 to 27 C calculated as below

Cp(HNO3)*(300-273) + N(No. of moles of H2O added)*Cp(H2O)*(300-273)

From the above table you can easily plot the data and the curve is like bell shape with weight % on x-axis and Enthalpy on y-axis

For the part b the question is not clear what is exactly happening

Enthalpy of reaction moles of H2O added Grams of H2O added Total Wt of solution Wt%ofHNO3inSolution 3350-3577.5=-227.5 0.1 1.8

64.8

97.22 5440-3780=1660 0.2 3.6 66.6 95.02 6900-3982.5=2917.5 0.3 5.4 68.4 92.1 8370-4387.5=3982.5 0.5 9 72 87.5 10880-4731.75=6148.25 0.67 12.06 75.06 83.93 14230-5400=8830 1 18 81 77.77 17150-6412.5=10737.5 1.5 27 90 70 20290-7425=12865 2 36 99 63.63 24060-9450=14610 3 54 117 53.84 25940-11475=14465 4 72 135 46.67 27820-13500=14320 5 90 153 41.17 30540-23625=6915 10 180 243 25.92 31170-43875=-12705 20 360 423 14.9

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