Calculate Hon in kJ/mol for the following reaction if 20.0 mL of a 1.50 M HCl(aq

Question

Calculate Hon in kJ/mol for the following reaction if 20.0 mL of a 1.50 M HCl(aq) is reacted with 15.00 mL of a 2.00 M NaOH(aq) in a coffee-cup calorimeter. The temperatures of the HCl solution and calorimeter before the solutions are mixed was 22.0°C, and the temperature of the NaOH solution was 21.8°C. The solutions were mixed and the temperature was monitored with the temperature reaching a maximum of 36.5°C. (The Ccal for the calorimeter was determined to be 8.75 J/°C. Assume the heat capacity and density of the solutions to be.that of water.) (0.5 pt)

Explanation / Answer

volume of HCl = 20 ml , density= 1 g/ml, mass of HCl= volume* density = 20*1= 20 gm

volume of NaOH= 15ml, density = 1g/ml, mass of NaOH= 15*1= 15 gm

heat added to NaOH due to exothermic reaction of NaOH and HCl= mass of HCl solution* specific heat* change in temperature= 15*4.184*(36.5-21.8) =923 joules

heat added to HCl due to the reaction = 20*4.184*(36.5-22.6)=1163 joules

heat added to calorimeter= heat capacity* change in temperature = 8.75*(36.5-22.6)= 122 joules

total heat due to the reaction = 923+1163+122 =2208 joules

mole of HCl= molarity* volume in L= 1.5*20/1000 = 0.03, mole of NaOH= 2*15/1000 =0.03

the reaction of HCl and NaOH is   NaOH+ HCl ------->NaCl+ H2O

molar ratio of HCl :NaOH= 1:!,

heat added per mole of HCl ( or NaOH)= 2208/0.03 J/mole =73600 j/mole

since there is a rise in temperature, enthalpy change is -ve. Enthalpy change/mole= -73600J/mole= -73.6 Kj/mole

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