Joules to kJ/mol conversion

Question

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.

Answer 1

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.