You have a 5.82 calorimeter containing some water at roomtemperature with 49.2mL

Question

You have a 5.82 calorimeter containing some water at roomtemperature with 49.2mL of 1.1M HCl. When you add 25.3mL of 1.9MNaOH, the temperature raises to 31.1deg C. The total mass of thesolution is 77.36g. (The conditions of the room have not changed.You may assume the specific heat of your solutions are the same ofthat of water.) 1a. Calculate the heat flow of the solution. 1b. Calculate the heat flow for the reaction. 1c. Calculate the molar enthalpy for the reaction (inkj/mol.). You have a 5.82 calorimeter containing some water at roomtemperature with 49.2mL of 1.1M HCl. When you add 25.3mL of 1.9MNaOH, the temperature raises to 31.1deg C. The total mass of thesolution is 77.36g. (The conditions of the room have not changed.You may assume the specific heat of your solutions are the same ofthat of water.) 1a. Calculate the heat flow of the solution. 1b. Calculate the heat flow for the reaction. 1c. Calculate the molar enthalpy for the reaction (inkj/mol.).

Explanation / Answer

First we assume that the room temperature is25oC Assume that no heat is lost to the surroundings,qsys = qaoln +qrxn   , So, qsoln =- qrxn Total mass of the solution given = 77.36 g Therefore heat of solution = qsoln = 77.36 g* 4.184 J/g.oC * (31.1 - 25)                                                     = 1974.41 J This is the heat flow of the solution, we know thatqsoln = - qrxn , the heat flow of thereaction is also -1974.41 J The number of moles of HCl present = 1.1 mo/L *0.0492 L= 0.05412 mol Number of moles of NaOH = 1.9 mol/L * 0.0253L = 0.04807 mol Thus we expect that there are 0.04807 mole of bothreactants participate in the neutralization reaction. Hence -                     The heat of neutralization = - 1974.41 J / 0.04807 mol                                                             = -41073.6 J/mole                                                             = -41.0736 kJ/mole        

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