Question

If 50.0 mL of 0.88 M H2O2 and 10.0 mL of 0.50 M Fe(NO3)3 were combined and a temperature change of -6.61°C was observed, The specific heat of water is 4.18 J/(g * ∘C)

Calculate the heat of reaction (in kilojoules). Record your answer with the proper significant figures and include the correct sign if needed.

Answer 1

Balanced equation:
3 H2O2 + 2 Fe(NO3)3 ====> 6 HNO3 + 2 FeO3

Since heat is absorbed it is endothermic reaction

we can use the following formula

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat (units J/g∙°C), ∆ is a symbol meaning "the change in"

∆T = change in temperature (°C Celcius)

Heat gained by reaction = Heat released by solution  

Heat released by solution   

Q = ? m = 50 ml + 10 ml = 60 ml = 60 gm c = 4.184  J/g∙ °C

∆T =6.61  °C

Q = 60 gm x 4.184 J/g∙°C x 6.61 °C =  1,659.37 Joules

Moles of H2O2 = 50 x 0.88 / 1000 = 0.044 Moles

Moles of Fe(NO3)3 = 10 x 0.5 /1000 = 0.005 Moles

Fe(NO3)3 is limiting regent

heat of the reaction   = 1659.37 J = 1.659 kJ

Enthalpy of dissolution per mole = 1.659 kJ x 2 / 0.005Mole = 663.6 kJ /Mole