Question

# A 50.0 mL sample of 0.300 M NaOH is mixed with a 50.0 mL sample of...

A 50.0 mL sample of 0.300 M NaOH is mixed with a 50.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH°rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2) that the density and the heat capacity of the resulting solution are the same as water.

Number of moles,n=molarity * volume in L

So number of moles of NaOH = number of moles of HNO3

= 0.300M*50.0mL*10^-3L/mL

=15*10^-3mol

Total volume of solution,V=50.0+50.0=100.0mL

Density of solution is 1g/mL

So mass of solution,M=volume*density

=100g

Dt=Change in temperature=37.00-35.00=2.00 oC

C= specific heat capacity =4.186 J/goC

So heat absorbed,q=mCDt =837.2 J

This is the heat energy released per 15*10^-3 mol of NaOH

So the heat energy released per mol of NaOH=837.2 x15*10^-3 =55.8*10^-3 J=55.8kJ

So Horxn=-55.8kJ

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