Question

What is the Amount of heat evolved when $500{\mathrm{cm}}^3$ of $0.2\mathrm{m}{\mathrm{H}}_2{\mathrm{SO}}_4$ is mixed with$200{\mathrm{cm}}^3$ of $0.5\mathrm{m}\mathrm{NaOH}$ solution?

Answer 1

Step 1: Mole calculation:

$$\begin{gathered} \mathrm{Moles}\mathrm{of}\mathrm{HCl}=\mathrm{m}\mathrm{x}\mathrm{v} \\ =500\mathrm{x}0.1 \\ \left[{\mathrm{H}}^{+}\right]=50\mathrm{milli}\mathrm{moles} \end{gathered}$$

$500{\mathrm{cm}}^3$ of $0.1\mathrm{M}\mathrm{HCl}$ corresponds to $0.5\times 0.1=0.05$ moles of hydrogen ions.

$$\begin{gathered} \mathrm{Moles}\mathrm{of}\mathrm{NaOH}=\mathrm{m}\mathrm{x}\mathrm{v} \\ =200\mathrm{x}0.1 \\ \left[{\mathrm{OH}}^{-}\right]=400\mathrm{milli}\mathrm{moles} \end{gathered}$$
$200{\mathrm{cm}}^3$ of $0.2\mathrm{M}\mathrm{NaOH}$ corresponds to$0.2\times 0.2=0.04$ moles of hydroxide ions.

Step 2: Neutralization reaction:

• sodium hydroxide is completely neutralize by $\mathrm{HCl}$

$$\begin{gathered} {\mathrm{H}}^{+}+{\mathrm{Cl}}^{-}\to {\mathrm{H}}_2\mathrm{O} \\ \mathrm{The}\mathrm{enthalphy}\mathrm{of}\mathrm{neutralization}\mathrm{for}\mathrm{water}\mathrm{is}, \\ ∆\mathrm{H}=-57.3\mathrm{KJ}{\mathrm{mol}}^{-1} \end{gathered}$$

Step 3: Calculating the heat evolved in the reaction:

• Here, the limiting reagent is sodium hydroxide as it contains lower concentration.

$$\begin{gathered} \mathrm{Heat}\mathrm{evolved}=57.3\mathrm{x}0.04 \\ =2.292\mathrm{KJ} \end{gathered}$$

Hence, the heat evolved is $2.292\mathrm{KJ}{\mathrm{mol}}^{-1}$.