CHEMISTRY EXPERIMENT HELP!?

Posted on 2016-03-10 18:49:00 in Chemistry by MANDY

I performed an experiment where I added 50.0mL of NaOH to 50.0mL of HCl to create a neutralization reaction. My initial temperatures of NaOH and HCl were both 22.25 C. The final temperature of the solution was 29.4 C. The densities of the solutions are 1.00g/mL and the specific heat of the salt solution is the same as that of water, 4.18J/g C. 
The q for my solution is: 
q=-(100.0g)(4.18J/g C)(29.4 C - 22.25 C) 
q= -2.99kJ 
I calculated my enthalpy of neutralization by dividing my q for my solution by the number of moles in the limiting reagent, NaOH. 
Enthalpy=-2.99kJ/1.25mol 
Enthalpy=-2.39kJ/mol 

The enthalpies of formation of elements in their standard states at 25 C are taken to be zero by conversion. 
The enthalpy of formation for a 1M solution of H+(aq) is chosen by definition to be 0. 
The enthalpy of formation for H2O(l) is -285.8kJ/mol 
Question: ****** 
1.Then I am asked to calculate the enthalpy change using my experimental data for 
OH-(aq) + H+(aq) -> H2O(l) 
2. Next I am asked to calculate the enthalpy of formation for (OH-(aq))*(Standard Enthalpy of H2O) 


I am unsure where to begin on these two questions! Please help!

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