Answer to Question #97764 in General Chemistry for Brittany Wallace

Question #97764
When a solid dissolves in water, heat may be evolved or absorbed. The heat of dissolution (dissolving) can be determined using a coffee cup calorimeter.

In the laboratory a general chemistry student finds that when 9.26 g of CsBr(s) are dissolved in 106.00 g of water, the temperature of the solution drops from 23.80 to 20.92 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.74 J/°C.

Based on the student's observation, calculate the enthalpy of dissolution of CsBr(s) in kJ/mol.

Assume the specific heat of the solution is equal to the specific heat of water.

ΔHdissolution = kJ/mol
1
Expert's answer
2019-11-01T05:44:50-0400

Mass of the solution is "(9.26+106)g=115.26g"

Specific heat capacity of the solution is equal to specific heat capacity of the water.

C= 4.186J/(g.°C)

Change in "Temp = T_f-T_i"

Tf=20.92°C

Ti=23.80°C

Change in "Temp =20.92-23.80=-2.88\u00b0C"

"Q_{solution}=mC\u00d7change in Temp"

"Q_{solution} =-115.26\u00d74.186\u00d72.88J"

"Q_{solution} =-1389.538J"

"Q_{reaction}=-( Q_{solution}+Q_{calorimeter})"

"Q_{ calorimeter}=z\u00d7change in Temp"

"z=heat capacity of calorimeter=1.74J\/\u00b0C"

"Q_{calorimeter}=-(1.74\u00d72.88)J"

"Q_{calorimeter}=-5.011J"

"Q_{reaction}=-(Q_{calorimeter}+Q_{solution})"

"Q_{reaction}=-(-1389.538+(-5.011))J"

"Q_{reaction}=1394.549J"

"\u0394H_{dissolution}=Q_{reaction}\/n"

"n = number of moles =9.26\/212.81=0.044moles"

"\u0394H_{dissolution}=(1394.549\/0.044)J\/mol"

"\u0394H_{dissolution}=31,694.2955J\/mol=31.694KJ\/mol"

(Answer)




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