0 Heat of Reaction for HCI (a)NaOH (a) Volume of 1.0 M HCI (ml) Volume of 1.0 M

Question

0 Heat of Reaction for HCI (a)NaOH (a) Volume of 1.0 M HCI (ml) Volume of 1.0 M NJOH (ml) Mass of Solution (a) use 1.00 g/m Ti solution (temp of HCI just before reaction) Ti calorimeter (same as Ti solution, above) Tf solution (highest temp obtained) Tf calorimeter (highest temp obtained) So Iml 23 23 2 20.1 solution ) q solution (mass of solution) x (4.184 l/ gC) x (Tr solution 2645 117 142.8 J q calorimeter) use qcalorimeter . (21C) x (Tfcalorimeter-T. q reaction (U) 2981 23 use q a calorimeter+ q solution +qreaction from the first law of thermodynamics Enthalpy of Reaction (4Hrxn): Obviously, the heat of reaction will vary depending on how much HCI undergoes neutralization. If you had used twice as much HCI, and twice as much NaOH, your va reaction would be twice as high. Hence, enthalpy of a reaction is reported as the heat of reaction per the molar quantities shown in the balanced reaction. In this case, q reaction if one mole of HCI had reacted with one mol of NaOH (U/mol). To get this value, divide your heat of reaction, J, by the mols of lue of q acid usecd mols of HCI reacted 50.0 ml of 1.00 M HCI AHran U/mol) AHrxn (kJ/mol) 1kJ 1000 Average AHrxn The literature value of AHrxn for the neutralization reaction is -56.2 kJ/mol. How close was your average value? Discuss possible reasons for differences

Explanation / Answer

The volume used of 1.0 M HCl is = 50.1mL.

Now, 1M HCl means

1000 mL of solution has moles = 1 mol of HCl

50.1 mL has moles = 1×50.1/1000 = 0.0501 mol.So,

#Moles of HCl reacted = 0.0501

Hrexn = q rexn = 2989.92 J.So,

0.050 mol HCl give the energy = 2989.98 J

1 mol give energy = 2989.92 J/0.0501 mol= 59679.042J/mol

So, Hrexn(J/mol) = 59679.042 J/mol

Hrexn(kJ/mol) = 59.679 kJ/mol

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