The molar enthalpy change for the neutralization reaction: HCl (a)NaOH (aq)->NaC
ID: 1038313 • Letter: T
Question
The molar enthalpy change for the neutralization reaction: HCl (a)NaOH (aq)->NaCI (aq)+H20 (l) is AH--57.3 kJ/mol. This is an exothermic reaction which means heat is released. The heat released by this reaction can be calculated with the equation q nAH, where n is the number of moles of the limiting reagent. The heat released by this reaction increases the temperature of the reaction solution (the surroundings). Imagine you mix 50.00 ml of 1.00 M HCI with to 50.00 ml 1.00 M NaOH in a beaker at 23°C. Estimate the final temperature of the solution after the reaction has gone to completion.You will need to make a few assumptions: Assume no heat is lost to the environment. Assume the density of the reaction solution is the same as water, 1.00 g/mL. Assume the specific heat capacity of the reaction solution is the same as of water, 4.184 J/g deg.Explanation / Answer
The balanced chemical equation for the reaction is given as
HCl (aq) + NaOH (aq) -------> NaCl (aq) + H2O (l)
As per the stoichiometric equation,
1 mole HCl = 1 mole NaOH.
Moles HCl = moles NaOH = (50.0 mL)*(1 L/1000 mL)*(1.00 M) = 0.05 mole.
Complete reaction occurs and hence, we have no limiting reactant. Therefore, we can take 0.05 mole HCl as the moles of reactant.
The heat released by the reaction is absorbed by water; hence q is positive and is equal to
q = n*?H = (0.05 mole)*(57.3 kJ/mol) (molar enthalpy of the reaction is given, i.e, heat released per mole of HCl)
= 2.865 kJ = (2.865 kJ)*(1000 J/1 kJ) = 2865 J.
We have (50.0 + 50.0) mL = 100.0 mL of reactants at an initial temperature of 23°C. The density of the mixture is 1.00 g/mL; therefore, the mass of the mixture is (100.0 mL)*(1.00 g/mL) = 100.0 g.
The specific heat capacity of the reaction mixture is 4.184 J/g.°C.
Since no heat is lost, we can write
2865 J = (mass of mixture)*(specific heat capacity of mixture)*(change in temperature of the mixture)
====> 2865 J = (100.0 g)*(4.184 J/g.°C)*(t – 23)°C where t°C = final temperature of the mixture
====> 2865 J = (418.4 J)*(t – 23)
====> t – 23 = (2865 J)/(418.4 J) = 6.8475
====> t = 6.8475 + 23 = 29.8475 ? 29.85
The final temperature of the mixture is 29.85°C (ans).
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