Question

If 15.0 mL of a 1.5M HCl solution at 21.50C is mixed with 25.0mL of a 1.5M NaOH solution at 21.50C that is in a calorimeter, and the final mixed solution temperature rises to28.50C, what is the ΔHrxnfor this process?Assume that Ccalorimeter= 35.5J/0C.

a.What is the balanced equation for the reaction?

b.What is the source of the heat that is causing the increase in temperature?

c.Calculate the amount of heat absorbed or lost for the HCl/NaOH solution. Assume aqueous conditions.(Cwater= 4.18 J/g·0C, d =1.0 g/mL)

d.Calculate the amount of heat absorbed or lost by the calorimeter. The calorimeter’s initial temperature is the same as the solution that is initially inside it.

e.Determine the amount of heat absorbed or lost during this reaction.

f.Which of the reactants is the limiting reagent? Determine the moles of the limiting reagent.

g.Determine the amount of heat given off per mole of limiting reagent(ΔHrxn).Make sure to include an appropriate sign indicating whether it is an endothermic (+) or exothermic (-) process. Provide your answer in units of kJ/mol.

Answer 1

If 15.0 mL of a 1.5M HCl solution at 21.50C is mixed with 25.0mL of a 1.5M NaOH solution at 21.50C that is in a calorimeter, and the final mixed solution temperature rises to28.50C, what is the ΔHrxnfor this process?Assume that Ccalorimeter= 35.5J/0C.

a.What is the balanced equation for the reaction?

HCl(aq) + NaOH (aq) ==> NaCl(aq) + H₂O (l)

b.What is the source of the heat that is causing the increase in temperature?

Neutralization reaction, which is an exothermic reaction.

H+(aq) + OH-(aq) ==> H₂O(l)

c.Calculate the amount of heat absorbed or lost for the HCl/NaOH solution. Assume aqueous conditions.(Cwater= 4.18 J/g·0C, d =1.0 g/mL)

Q = m*Cwater* dT = ( 40 g*4.18 J/g·0C* 7C) = 1170.4 J

d.Calculate the amount of heat absorbed or lost by the calorimeter. The calorimeter’s initial temperature is the same as the solution that is initially inside it.

Ccal*dT = ( 35.5 J/C *7C) = 248.5 J

e.Determine the amount of heat absorbed or lost during this reaction.

It and exothermic reaction :

Q (heat lost) =   1418.9 J

f.Which of the reactants is the limiting reagent? Determine the moles of the limiting reagent.

HCl is a limiting reagent.

moles of the limiting reagent = 0.015 L *1.5 mol/L = 0.0225 moles

g.Determine the amount of heat given off per mole of limiting reagent(ΔHrxn).Make sure to include an appropriate sign indicating whether it is an endothermic (+) or exothermic (-) process. Provide your answer in units of kJ/mol.

It is an exothermic (-) process

ΔHrxn = - ( 1418.9 J / 0.0225 moles) = - 63.06 kJ/mol