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Ammonia (NH3) boils at -33.4 oC. It has an enthalpy of vaporization of 23.35 kJ/mol. The...

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Chemistry

Ammonia (NH3) boils at -33.4 oC. It has anenthalpy of vaporization of 23.35 kJ/mol. The liquid has a specificheat of 80.8 J/mol oC, and the gas has a specific heatof 35.06 J/mol 0C.

A 34.06 gram sample of ammonia was heated, at 1 atmospherepressure, from -40 oC to room temperature (25oC). How many kJ of thermal energy did thesample absorb from the surroundings in this process?

I got 101263.8J (101.3kJ) and it is NOT correct, pleasehelp with the correct answer.

Answer & Explanation Solved by verified expert
4.2 Ratings (899 Votes)

Mass of amonia = 34.06 gm
Molar mass = 17.03 gm
Number of moles, n = mass/ molar mass = 34.06/17.03 = 2 mol

Step 1: Raise temperature of liquid amonia from -40 oC to -33.4 oC
Q1= n*C*(Tf - Ti)
    = 2*80.8*(-33.4 - (-40))
    =1066.56 J

Step 2: Vapourise the liquid ammonia at -33.4 oC
Q2= n*C
       = 2*23.35 KJ
       = 76.7 KJ
       = 76700 J

Step 3: Raise temperature of vapour amonia from -33.4 oC to 25 oC
Q3= n*C*(Tf - Ti)
    = 2*35.06*(25-(-33.4))
    =4095.01 J

Total heat absorbed = Q1+Q2+Q3
                                          = 1066.56 + 76700 + 4095.01
                                          =81861.57 J
                                           = 81.86 KJ
Answer:   81.86 KJ
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