When calculating the thermal energy released during an acid-base neutralisation using the equation $Q = mc\,\Delta T$, and then using $\Delta H = - Q/n$ to calculate the enthalpy change of neutralisation, why is the amount of substance of one reactant used ($n$ in the second equation) rather than the total amount of substance of acid and base reacting together?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Consider the following reaction ($\Delta H_{\text{rxn}}$ is the enthalpy change for one mole of the reaction, or the amount of moles of each substance found in a balanced equation: in this case, 1 mole $\ce{A}$, 1 mole $\ce{B}$):
$$\ce{A->B}\qquad\Delta H_{\text{rxn}}=~?$$
You are given 5 moles of $\ce{A}$, and you measure the change in heat to be $\pu{-100 kJ}$. However, this $\pu{-100 kJ}$ is actually just the $\Delta H$ of this reaction: $$\ce{5A->5B}\qquad\Delta H=\pu{-100~kJ}$$
Looks the same, right? That's because it is the same. However, if you want to calculate the enthalpy change of the first reaction, you are going to have to divide the measured enthalpy change, $\pu{-100 kJ}$, by 5. Thus, the $\Delta H_{\text{rxn}}=\pu{-20kJ}$.
Now consider this: $$\ce{A +B->C}\qquad\Delta H_{\text{rxn}}=~?$$ If you were again given 5 moles of $\ce{A}$, along with 5 moles of $\ce{B}$, and you measured the $\Delta H$ to be $\pu{-100 kJ}$, you would get something similar to the first scenario: $$\ce{5A + 5B -> 5C}\qquad\Delta H=\pu{100kJ}$$
Looks familiar right? Now, if you wanted to calculate the $\Delta H_{\text{rxn}}$, you wouldn't divide 100 by 10; you would divide it by 5, since the second reaction is just $5x$ the first one.
Side-note: This same reaction would be written if we have more than 5 moles of $\ce{B}$. Since $\ce{A}$ and $\ce{B}$ react in a 1 : 1 ratio, adding more $\ce{B}$ than $\ce{A}$ would mean that there is excess $\ce{B}$.
