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Calculate the enthalpy of netralization per mole of NH4Cl formed in kJ in the following...

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boblee1245 | Student | eNoter

Posted February 19, 2012 at 9:58 PM via web

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Calculate the enthalpy of netralization per mole of NH4Cl formed in kJ in the following case:

A student mixed 50 ml of 1 M HCl at 20.5 C with 50 ml of 1 M NH3 at 20.5 C. After mixing, the temperature rose to 23.2 C. The heat capacity has been determined to be 18.0 J\g*C.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 19, 2012 at 10:30 PM (Answer #1)

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A student mixes 50.0 ml of 1 M HCl at 20.5 degrees Celsius with 50.00 ml of 1 M NH3 at 20.5 C. This results in the temperature rising to 23.2 C.

The rise in temperature is 23.2 - 20.5 = 2.7 C

The equation of the chemical reaction between HCl and NH3 is:

NH3 + HCl --> NH4Cl

50 mL each of 1 M HCl and 1 M NH3 solutions take part in the neutralization reaction. The number of moles of NH4Cl formed is 50*10^-3

It is assumed the heat capacity is of the solution of compounds present initially as well as what is formed finally. And this is equal to 18 J/g*C with the density of the solutions approximately 1g/ml

As the temperature of 100 g of solution rises by 2.7 C when 50*10^-3 mole of NH4Cl is formed, the heat required is 4860 J. In the reaction 0.05 mole of NH4Cl is formed.

This gives the enthalpy of neutralization as is 4860/50*10^-3 = 97.2 kJ/mole

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