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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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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