Question

25 mL of 1.0 M NaOH(aq) is placed in the coffee cup calorimeter and measure the temperature as 18.0 0C. Addition of 25 mL of 1.0 M HCl causes the temperature to rise to 25.0 0C. What is the Hrxn for this reaction of neutralization of NaOH? (dH2O = 1.00 g/mL and cs, H2O = 4.18 J/g0C))

Answer 1

Moles of NaOH= molarity* Volume= 1*25/1000=0.025 moles

Moles of HCl added = 1*25/1000 =0.025 moles

The reaction is HCl+ NaOH---> NaCl+H2O

1 mole of HCl requires 1 moles of NaOH

Volume of solution after mixing = 25+25= 50ml

Density =1g/ml, mass of the solution =25*1 =25gm

Specifc heat = 4.18 j/g.deg.c

temperature rise = 25-18= 7 deg.c

Heat of reaction= mass*speciifc heat* temperature difference= 50*4.18*7=1463 joules

0.025 moles of NaOH addition resulted in 1463 joules of heat

for 1 moles =1463/0.025=58520 joules/moles= 58.520 Kj/mol