A 50.00 mL sample of 0.5012 M KOH at 21.3 degree C was added to a calorimeter co

Question

A 50.00 mL sample of 0.5012 M KOH at 21.3 degree C was added to a calorimeter containing 50.00 mL of 0.4998 M HCI at 22.4 degree C. (The densities of these diluted solutions can be determined by using the appropriate temperatures and densities located in Table 10.1. Note that the units for density in those tables are in kg/m^3. Convert your densities to g/mL before use!) The maximum temperature, 25.2 degree C, of the resulting reaction was determined graphically. The heat capacity of the calorimeter was 46.19 J middot degree C^-1. For the following reaction, calculate the indicated values. HCl_(aq) + KOH_(aq) rightarrow KCl_(aq) + H_2 O_(I)

Explanation / Answer

Moles of HCl= molaity* Volume in L= 0.4998*50/1000=0.02499 moles

Mass of HCl= moles* Molecular weight of HCl= 0.02499*36.5gm=0.9121 gms

Moles of KOH= 0.5012*50/1000=0.02056

Mass of KOH= 0.02056*56=1.40 gms

Mass of mixture= 0.9121+1.4=2.3121 gms

Temperature difference for KOH= 25.2-21.3= 3.9 deg.c and for HCl= 25.2-22.4= 2.8 deg.c

Assuming LR is product, moles of KCl=0.02499 ( limting reactant). Mass of KCl =0.02499*74.5=1.86 gms

Since density data is not available, the solution is assumed to be dilute and volume of mixed solution= 100ml density =1 g/cc ( assumed).

Heat given to solution =50*1*4.184*(25.2-21.3)+ 50*4.184*(25.2-22.4)= 1360 Joule

Heat taken by Calorimeter= 46.19*(25.2- 21.3)=176 Joules

This heat is not available as it is taken by Calorimeter

Hence Enthalpy change= 1360-176=1184 Joiules

Hence enthalpy change/ mole= 1184/0.02499=47379 Joules/mole= 47.38 Kj/mole

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