Molar Enthalpy, Grade 12 Chemistry?

S

Scott

New member
In a student lab, 60.0mL of 0.700mol/L sodium hydroxide solution was neutralized with 40.0mL of excess sulfuric acid solution. The temperature increased by 5.6C

I know that you would end up using nHx=mct but its the double volume that is screwing me up.

Answer in back of text is -5.57 kJ/mol
 
Let the NaOH solution be called NS. Let the total solution be called TS and assume its density is 1.00g/1.00mL..

60.0mLNS x 0.700molNaOH/1000mLNS = 0.042 mole NaOH

100.0gTS x 5.6degC x -4.184J/g-degC = -2343 J

-2343J/0.042mol = -55,787 J/mole = -55.7 kJ/mole
 
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