Joules to kJ/mol conversion

In my AP Chem class we are working on testing Hess's Law and conducted three reactions. Then enthalpy changes of the 1) $\ce{NaOH + HCl}$ and 2) $\ce{NaOH + NH4Cl}$ to predict the enthalpy change of 3) $\ce{HCl + NH3}$.


I used the equation $q = C\_pm\Delta T$. We determined the $q$-values using a method given to us by our teacher. She told us to assume a denisty of $1.03\ \mathrm{g/mL}$ for all solutions, we then multiplied the density by $50\ \mathrm{mL}$ to determine the grams of each reactant present. That value is the $m$ term. We used $4.18\ \mathrm{J/(g\ ^\circ C)}$ for the specific heat of water. The temperature change for the first reaction was $14.7\ \mathrm{^\circ C}$, the second was $1.2\ \mathrm{^\circ C}$, and the third was $9.8\ \mathrm{^\circ C}$.


The first reaction yielded a q value of $6328.938\ \mathrm J$, the second yielded $516.648\ \mathrm J$, and the third yielded $4219.292\ \mathrm J$.


The next step would be to convert this $q$ values into $\mathrm{kJ/mol}$. I consulted my lab partner who did the following operation:


$$\mathrm{kJ/mol} = 6328.938\ \mathrm J \times 1\ \mathrm{kJ}/1000\ \mathrm J \times x/0.2\ \mathrm{mol}$$


In his math, there was no numerator on the $0.2\ \mathrm{mol}$ term. The mol value he used came from there being 0.2 moles of the reactants present. I believe his math to be incorrect, but as my teacher is out sick and there are no other chemistry teachers present, I cannot determine the correct conversion. I believe it to be:


$$\mathrm{kJ/mol} = 6328.938\ \mathrm J \times 1\ \mathrm{kJ}/1000\ \mathrm J \times 6.022^{23}/0.2\ \mathrm{mol}$$


Both of the conversions yield values with significant % error. Any assistance is appreciated.




Accepted book enthalpies are in the units $\mathrm{kJ/mol}$. 


Simply convert $\mathrm J$ to $\mathrm{kJ}$ by dividing by $1000$. Since $1\ \mathrm{kJ} = 1000~\mathrm{J}$. 


The moles of dilute aquaous solutions is best determined by volume (L) times concentration (mol/L).$$n = V \times c$$ You would use the *moles* of the *product*. If there is, in fact, "$0.2\ \mathrm{mol}$ of the reactants present" then there would be $0.2\ \mathrm{mol}$ of the products present. since the chemical reactions are $\ce{1 mol + 1 mol -> 1 mol + 1 mol}$. (You are not interested in the number of atoms per mole, so you would not use Avogadro's number.)


$$q= C\_\mathrm{sol} m\Delta T$$ where $C\_\mathrm{sol}= 4.06~\mathrm{J/(g \ ^\circ C)}$ (Specific heat capacity for dilute aqueous solutions)


$$q= C\_\mathrm{sol} m\Delta T / n\_\mathrm{product}$$ ($n = \text{moles product}$)


Remember that Hess's Law states that 



> 
> if a reaction can be expressed as the sum of two or more reactions, then the heat flow for the overall reaction is equal to the sum of the heat flows.
> 
> 
> 


Hence, the sum of the first 2 reactions should be equal to the 3rd reaction. 


1. $\displaystyle \ce{NaOH + HCl -> NaCl + H2O}$
2. $\displaystyle \ce{NH4OH + NaCl -> NaOH + NH4Cl}$ (Reversed and Enthalpy sign changed)
3. $\displaystyle\ce{HCl + NH4OH -> NH4Cl + H2O}$


Note how $\ce{NaOH}$ and $\ce{NaCl}$ cross out leaving the third reaction. 


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