a) When a 15.5 g sample of potassium chlorate dissolves in 100.0g of water in a

Question

a) When a 15.5 g sample of potassium chlorate dissolves in 100.0g of water in a calorimeter, the temperature of the water/solution drops from 25.0 degrees celcius to 13.7 degrees celcius. Write the balanced equation for this solution process.

b)Calculate the enthalpy change (in kJ/mol) for the solution process of part (a) (assuming the specific heat of the solution is essentially the same as the solvent)

c) is the process endothermic or exothermic?

d)What is the amount of heat change of the salt (system)?

e)What is the amount of heat change for the solvent (surroundings)?

Explanation / Answer

KClO3-------> K+ ClO3-   delH= +Ve

b) Mass of mixture= 100+15.5= 115.5 gm. Assumig specific heat of mixture same as that of water, specific heat of water= 4.18 j/g.deg.c and

enthalpy of mixing = mass* specific heat* temperature difference= 115.5*4.18*(13.7-25)=-5431.91j/

This is enthalpy change for 15.5 gms of KClO3, moles of KClO3= 15/122.5 ( Mass/Molecular weight)= 0.122 moles

enthalpy change of mixture= 5431.91/0.122=-44360.6 J/mole= -44.360 Kj/Mole

since there is a temperature drop, the reaction is endothermic

d) heat change in salt= 15*specific heat of KClO3* temperature difference= 15*0.81*(13.7-25)=-137.3 Joules

heat change in water= 100*4.18*(13.7-25)=-4723.4 Joles

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