Animated Solution for Chemistry - Chemical Thermodynamics: 200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralisation of this reaction is $-57.1\text{ kJ}$. The increase in temperature in ${}^{\circ }\text{C}$ of the system on mixing is $x\times 10^{-2}$. The value of $x$ is ......... (Nearest integer) [Given, specific heat of water = $4.18{\text{ J g}}^{-1}{\text{ K}}^{-1}$ Density of water = $1.00{\text{ g cm}}^{-3}$] (Assume no volume change on mixing)

A Tale of Two Solutions

Imagine you are standing in a laboratory, holding two beakers. In one hand, you have $200\text{ mL}$ of $0.2\text{ M}$ Hydrochloric Acid ($\text{HCl}$), a strong acid. In the other, you hold $300\text{ mL}$ of $0.1\text{ M}$ Sodium Hydroxide ($\text{NaOH}$), a strong base. When you mix them, a classic neutralization reaction occurs, forming salt and water. But this isn't just a chemical transformation; it's a thermal event. The reaction releases heat, and our goal is to figure out exactly how much the temperature of the resulting solution will rise.

Before we can calculate the heat released, we must determine how much water is actually formed. This requires us to find the limiting reagent—the reactant that will be completely consumed first. We do this by calculating the millimoles of each reactant.

For $\text{HCl}$, the millimoles are calculated as:

For $\text{NaOH}$, the calculation is:

Comparing the two, we see that we have fewer millimoles of $\text{NaOH}$. Therefore, $\text{NaOH}$ is our limiting reagent. The reaction will stop once all $30\text{ mmol}$ of $\text{NaOH}$ are consumed, meaning exactly $30\text{ mmol}$ (or $30\times 10^{-3}\text{ mol}$) of water will be produced.

We are given that the molar heat of neutralization is $-57.1\text{ kJ/mol}$. The negative sign simply indicates that the process is exothermic—heat is released into the surroundings. To find the total heat energy ($q$) released by our specific reaction, we multiply the moles of water formed by the magnitude of this molar heat.

Substituting our values, and remembering to convert kilojoules to joules to match our specific heat units later:

The $1000$s cancel out beautifully, leaving us with:

This $1713\text{ J}$ is the total thermal energy injected into the solution.

Now, how does this energy translate into a temperature rise? We use the fundamental calorimetry equation:

We need the total mass ($m$) of the solution. The total volume is the sum of the two individual volumes: $200\text{ mL}+300\text{ mL}=500\text{ mL}$. Given the density of water is $1.00{\text{ g/cm}}^3$ (which is equivalent to $1.00\text{ g/mL}$), the mass of our solution is exactly $500\text{ g}$.

Rearranging our equation to solve for the change in temperature ($\Delta T$):

Calculating the denominator gives us $2090$. Dividing $1713$ by $2090$ yields approximately $0.8196\text{ K}$. Since a change of one Kelvin is exactly equal to a change of one degree Celsius, our temperature rise is $0.8196^{\circ }\text{C}$.

The problem asks for the answer in the specific format of $x\times 10^{-2}$. We can rewrite our result to match this:

Rounding $81.96$ to the nearest integer gives us $82$. Thus, the value of $x$ is $82$.