# Chemistry 12: Calculating changes in enthalpy [closed]

I was wondering if I could please get some help with this:

In a coffee-cup calorimeter $$\pu{100.0 mL}$$ of $$\pu{1.0 M}\ \ce{NaOH}$$ and $$\pu{100.0 mL}$$ of $$\pu{1.0 M}\ \ce{HCl}$$ are mixed. Both solutions are originally at $$\pu{24.6 ^\circ C}$$. After the reaction, the temperature is $$\pu{31.3 ^\circ C}$$. Assuming all solutions have a density of $$\pu{1.0 g/cm^3}$$ and a specific heat capacity of $$\pu{4.181 J/(g ^\circ C)}$$. What is the enthalpy change for the neutralization of $$\ce{HCl}$$ by $$\ce{NaOH}$$?"

This is what I did, but I'm not sure if it's correct:

\begin{align} q(\text{surroundings}) &= m \cdot c \cdot \Delta T\\ &= \pu{200 g} \cdot \pu{4.184 J/(g ^\circ C)} \cdot (\pu{31.3 ^\circ C} - \pu{24.6 ^\circ C})\\ &= \pu{5.61 kJ}\\[2em] q(\text{system}) - \pu{5.61 kJ} &= n \cdot \Delta H - \pu{5.61 kJ}\\ &= \pu{0.1 mol} \cdot \Delta H \end{align}

Thus: $$\Delta H = \pu{-56.1 kJ/mol}$$

In general, homework is not welcome here in the forum, which is probably why your question was voted down. However, since you have already given a solution here and only ask for a confirmation, I want to be so nice and reply to your question.

Thus: $$Q(\text{neutralization}) + Q(\text{warm the solution}) + Q(\text{warm the calorimeter}) = 0$$
In your example, it is assumed for simplicity that the calorimeter does not consume heat to heat up. Therefore $$Q(\text{warm the calorimeter}) = 0$$.
\begin{align} Q(\text{neutralization}) + Q(\text{warm the solution}) &= 0\\ n \Delta H + m c \Delta T &= 0\\ n \Delta H &= - m c \Delta T\\ \Delta H &= -\frac{m c \Delta T}{n}\\ &= \frac{\pu{200 g} \cdot \pu{4.184 J/(g ^\circ C)} \cdot (\pu{31.3 ^\circ C} - \pu{24.6 ^\circ C})}{\pu{0.1 mol}}\\ &= \pu{-56.1 kJ/mol} \end{align}