Problem 99 - Chemistry An Atoms First Approach

Question

The enthalpy of neutralization for the reaction of a strong acid with a strong base is \(-56 \mathrm{kJ} / \mathrm{mol}\) water produced. How much energy will be released when \(200.0 \mathrm{mL}\) of \(0.400 \mathrm{M} \mathrm{HNO}_{3}\) is mixed with \(150.0 \mathrm{mL}\) of \(0.500 \mathrm{M}\) KOH?

Step-by-Step Solution

Answer

When 200.0 mL of 0.400 M HNO₃ is mixed with 150.0 mL of 0.500 M KOH, the limiting reactant is KOH, and 0.075 mol of water is produced. The energy released during the reaction is \(0.075 \mathrm{mol} \times (-56 \mathrm{kJ/mol}) = -4.2 \mathrm{kJ}\). Thus, 4.2 kJ of energy is released.

Step 1: Calculate the moles of the reactants

We are given the volume and molarity of both HNO₃ and KOH. We can use the formula n = M × V to find the moles of the reactants. For HNO₃: n(HNO₃) = 0.400 M × 200.0 mL = 0.400 mol/L × 0.200 L = 0.080 mol For KOH: n(KOH) = 0.500 M × 150.0 mL = 0.500 mol/L × 0.150 L = 0.075 mol

Step 2: Determine the limiting reactant

Since the reaction involves the combination of an acid (HNO₃) and a base (KOH), the balanced chemical equation looks like this: HNO₃ + KOH → KNO₃ + H₂O From the balanced equation, we can see that the ratio of moles between HNO₃ and KOH is 1:1. We need to determine which one of the reactants is the limiting reactant by comparing their mole ratio. Since 0.080 mol of HNO₃ reacts with 0.075 mol of KOH (1:1 ratio), HNO₃ is present in excess, and KOH is the limiting reactant.

Step 3: Calculate the moles of water produced

The limiting reactant determines the moles of water produced. In this case, since KOH is the limiting reactant, 0.075 mol of KOH produces 0.075 mol of H₂O.

Step 4: Calculate the energy released

Now that we have determined the number of moles of water produced, we can calculate the energy released. We know that the enthalpy of neutralization is -56 kJ/mol of water. Therefore, we can use the following formula: Energy released = moles of water × enthalpy of neutralization per mole of water Energy released = 0.075 mol × (-56 kJ/mol) = -4.2 kJ Since the energy released is negative, this tells us that 4.2 kJ of energy is released during the reaction.

Key Concepts

Energy Released in Reactions

Chemical reactions often involve the transfer of energy, particularly in forms like heat. During a reaction between a strong acid and a strong base, one such energy transfer takes place, specifically known as the enthalpy of neutralization. This is the energy change when an acid and a base react to form water and a salt.

In the case of the reaction between \( ext{HNO}_3\) and \( ext{KOH}\), energy is released as these compounds form water (\( ext{H}_2 ext{O}\)) and a salt (\( ext{KNO}_3\)). The enthalpy of neutralization for this reaction is \-56 \, ext{kJ/mol}\, meaning that for each mole of water produced, 56 kJ of energy is released. This released energy can sometimes be observed as a temperature increase in the solution. This is why understanding this energy change is crucial, especially when scale-up reactions are considered, as it helps predict temperature changes and ensures safe handling of reactions.

Always remember that the sign of the enthalpy change tells us about the nature of the reaction in terms of energy. A negative enthalpy of neutralization indicates that the reaction releases energy, i.e., it is exothermic.

Limiting Reactant

In a chemical reaction, knowing which reactant will run out first is important. This reactant is called the limiting reactant. It is the substance that determines the maximum amount of product that can be formed, thus "limiting" the reaction.

For our reaction, when \( ext{HNO}_3\) reacts with \( ext{KOH}\), the balanced equation is \( ext{HNO}_3 + ext{KOH} \rightarrow ext{KNO}_3 + ext{H}_2 ext{O}\). The mole ratio between \( ext{HNO}_3\) and \( ext{KOH}\) is 1:1 according to the balanced equation. Given the initial moles calculated: 0.080 mol of \( ext{HNO}_3\) and 0.075 mol of \( ext{KOH}\), we can identify \( ext{KOH}\) as the limiting reactant. This is because it has fewer moles compared with \( ext{HNO}_3\) in the required 1:1 ratio.

Once \( ext{KOH}\) is completely used up, no more water can be produced, which makes \( ext{KOH}\) the limiting reactant. Knowing the limiting reactant helps predict exactly how much product can be formed and how much energy will be released, as it directly affects these calculations.

Moles and Molarity Calculation

To perform many calculations in chemistry, understanding moles and molarity is fundamental. Moles represent an amount of substance. It's a very large number of entities—like atoms or molecules. Molarity, on the other hand, is a way of expressing the concentration of a solution. It is defined as the number of moles of a solute divided by the volume of the solution in liters.

Here's how you can calculate moles from molarity: Use the formula \(n = M \times V\), where \(n\) is the number of moles, \(M\) is the molarity (in mol/L), and \(V\) is the volume of the solution (in liters). For \( ext{HNO}_3\) in our exercise, the molarity was 0.400 M, and the volume was converted from 200.0 mL to 0.200 L (since 1 L = 1000 mL). Thus, \(n( ext{HNO}_3) = 0.400 \, ext{mol/L} \times 0.200 \, ext{L} = 0.080 \, ext{mol}\).

Similarly, calculate for \( ext{KOH}\) using its molarity and volume. These steps allow you to determine how much of each reactant you have, which is vital for identifying the limiting reactant and the amount of product formed. Understanding how to calculate moles and molarity is an essential skill for any chemistry student and is a powerful tool in both academic and real-world chemical applications.