Problem 20 - Chemistry in 30 days

Question

The heat of neutralisation of strong base and strong acid is \(57.0 \mathrm{~kJ}\); the heat released when \(0.5 \mathrm{~mol}\) of \(\mathrm{HNO}_{3}\) solution is added to \(0.20 \mathrm{~mol}\) of \(\mathrm{NaOH}\) solution is (a) \(57.0 \mathrm{~kJ}\) (b) \(28.5 \mathrm{~kJ}\) (c) \(11.40 \mathrm{~kJ}\) (d) \(34.9 \mathrm{~kJ}\)

Step-by-Step Solution

Answer

The heat released is 11.4 kJ; option (c) is correct.

Step 1: Understanding the Problem

The heat of neutralization for a strong acid and a strong base reaction is given as \(57.0 \text{ kJ/mol}\). We are asked to find the heat released when 0.5 moles of \(\text{HNO}_3\) reacts with 0.2 moles of \(\text{NaOH}\). Since \(\text{NaOH}\) is the limiting reagent, we will calculate the heat released using its amount (0.2 moles).

Step 2: Identifying the Limiting Reagent

First, we determine which reagent is limiting. We have 0.5 moles of \(\text{HNO}_3\) and 0.2 moles of \(\text{NaOH}\). Since each mole of \(\text{HNO}_3\) reacts with one mole of \(\text{NaOH}\), \(\text{NaOH}\) is the limiting reagent, as it is present in a smaller amount (0.2 moles vs. 0.5 moles).

Step 3: Applying Heat of Neutralization

Use the heat of neutralization value, which states that the reaction of 1 mole of a strong acid with 1 mole of a strong base releases \(57.0 \text{ kJ}\). Since \(0.2\) moles of \(\text{NaOH}\) will react with \(0.2\) moles of \(\text{HNO}_3\), only 0.2 moles of the reaction will occur.

Step 4: Calculating the Heat Released

Multiply the amount of the limiting reagent (0.2 moles) by the heat of neutralization per mole (57.0 kJ/mol): \[ \text{Heat released} = 0.2 \text{ mol} \times 57.0 \text{ kJ/mol} = 11.4 \text{ kJ} \]

Step 5: Finding the Correct Answer

According to our calculation, the heat released when 0.5 moles of \(\text{HNO}_3\) reacts with 0.2 moles of \(\text{NaOH}\) is \(11.4 \text{ kJ}\). Thus, the correct answer is option (c).

Key Concepts

Limiting Reagent

In a chemical reaction, the limiting reagent is the substance that is completely consumed first, determining the maximum amount of product that can be formed. In the example of mixing \(0.5\, \text{mol}\) of \(\text{HNO}_3\) and \(0.2\, \text{mol}\) of \(\text{NaOH}\), we need to establish which of these two is the limiting reagent.
The concept of the limiting reagent is crucial because once it is used up, the reaction cannot proceed any further. For a reaction between a strong acid and a strong base, such as \(\text{HNO}_3\) and \(\text{NaOH}\), the stoichiometry is 1:1. This means one mole of \(\text{HNO}_3\) reacts with one mole of \(\text{NaOH}\).
- **Assessment:** - \(\text{NaOH}\): **0.2 mol available** - \(\text{HNO}_3\): **0.5 mol available**
Considering the stoichiometric ratio, \(\text{NaOH}\) is clearly the limiting reagent because we only have 0.2 moles of it compared to 0.5 moles of \(\text{HNO}_3\). This results in the reaction stopping once all \(\text{NaOH}\) is consumed.

Strong Acid and Strong Base Reaction

A strong acid and strong base reaction is a type of chemical reaction known as neutralization. In this process, an acid and a base interact to produce water and a salt. This type of reaction is quite exothermic, meaning it releases a noticeable amount of energy in the form of heat.
Key characteristics include:- The reaction is typically rapid and complete, as both reactants dissociate completely in their solutions.
- Strong acids, like \(\text{HNO}_3\), and strong bases, such as \(\text{NaOH}\), disassociate to release \(\text{H}^+\) and \(\text{OH}^-\) ions, respectively. When these ions meet, they form water: \[ \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \]- As a result of this meeting, there’s a significant release of energy, recorded as the heat of neutralization.
In the given example, when mixing \(0.2\, \text{mol}\) of \(\text{NaOH}\) with \(0.5\, \text{mol}\) of \(\text{HNO}_3\), only \(0.2\) mol of reaction occurs due to the limiting nature of the base.

Enthalpy Change Calculation

The enthalpy change, specifically the heat of neutralization, gives you the amount of heat that is either absorbed or released during a reaction at constant pressure. It's expressed in kilojoules per mole (\(\text{kJ/mol}\)).
For neutralization reactions, where a strong acid reacts with a strong base, the heat of neutralization is usually around \(57.0 \text{ kJ/mol}\). This means that for every mole of acid reacting with a mole of base, \(57.0\,\text{kJ}\) of heat is released.
To calculate the total heat released during the reaction:- Determine the moles of the limiting reagent: In this case, \(\text{NaOH}\) is the limiting reagent with \(0.2\, \text{mol}\).- Multiply the moles of the limiting reagent by the heat of neutralization: \[ \text{Heat released} = 0.2 \text{ mol} \times 57.0 \text{ kJ/mol} = 11.4 \text{ kJ} \]This calculation shows that \(11.4\,\text{kJ}\) of heat is released when \(0.2\,\text{mol}\) of \(\text{NaOH}\) neutralizes with an equivalent amount of \(\text{HNO}_3\). Understanding enthalpy changes helps in predicting how energy dynamics influence chemical processes.