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 \(0.500 M \mathrm{KOH}\) ?

Short answer

The amount of energy released when $200.0 \mathrm{~mL}$ of $0.400 \mathrm{M} \mathrm{HNO}_{3}$ is mixed with $150.0 \mathrm{~mL}$ of $0.500 \mathrm{M} \mathrm{KOH}$ is $-4.2 \mathrm{~kJ}$.

Step-by-step solution

Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction of nitric acid (HNO3) with potassium hydroxide (KOH) is: \( HNO_3 + KOH \rightarrow KNO_3 + H_2O \)

Calculate the number of moles of HNO3 and KOH

Given the volume and molar concentration of HNO3 and KOH, we can calculate the number of moles of each reactant: Moles of HNO3 = (volume of HNO3) × (molar concentration of HNO3) = (200.0 mL) × (0.400 mol/L) = 0.200 L × 0.400 mol/L = 0.080 mol Moles of KOH = (volume of KOH) × (molar concentration of KOH) = (150.0 mL) × (0.500 mol/L) = 0.150 L × 0.500 mol/L = 0.075 mol

Determine the limiting reactant and calculate the moles of water produced

Now, we will compare the mole ratio of HNO3 and KOH to determine the limiting reactant, and use that to calculate the moles of water produced. Mole ratio of HNO3 to KOH = (moles of HNO3) / (moles of KOH) = 0.080 mol / 0.075 mol = 1.067 Since the mole ratio of HNO3 to KOH in the balanced chemical equation is 1:1, we can see that KOH is the limiting reactant. Therefore, the moles of water produced will equal the moles of KOH: Moles of water produced = moles of KOH = 0.075 mol

Calculate the energy released using the enthalpy of neutralization

We are given the enthalpy of neutralization per mole of water produced, which is -56 kJ/mol. To calculate the energy released, we will multiply this value by the moles of water produced: Energy released = (enthalpy of neutralization) × (moles of water produced) = (-56 kJ/mol) × (0.075 mol) = -4.2 kJ Thus, the amount of energy released when 200.0 mL of 0.400 M HNO3 is mixed with 150.0 mL of 0.500 M KOH is -4.2 kJ.

Key concepts

Limiting Reactant

In chemical reactions, the concept of a limiting reactant is crucial for determining how much product can be formed. The limiting reactant is the reactant that is completely consumed first, stopping the reaction from continuing and thus limiting the amount of products that can be formed. In the exercise we have, we mix nitric acid ( HNO_3 ) with potassium hydroxide ( KOH ). Based on the given volumes and molarities, we have calculated that HNO_3 presents in 0.080 moles, while KOH is present in 0.075 moles.

Since both reactants react in a 1:1 ratio, KOH , which has fewer moles initially, becomes the limiting reactant. Once all the KOH is used up, the reaction comes to a halt. Always identify the limiting reactant to know which reactant will dictate the maximum amount of product formed in a reaction.

Chemical Equation Balancing

Balancing chemical equations is an essential step in solving reaction problems. For our problem, we have the reaction of HNO_3 with KOH, which can be written as:\[HNO_3 + KOH \rightarrow KNO_3 + H_2O\]This equation is already balanced as there is the same number of each type of atom on both sides of the equation.

Balancing ensures conservation of mass and allows us to use the stoichiometry of the equation to relate the moles of reactants and products. Ensure every equation is balanced before proceeding with calculations, as an unbalanced equation can lead to incorrect conclusions about reactant or product quantities.

Energy Calculation

Energy calculations in chemical reactions often involve the enthalpy change, especially in reactions like neutralization. In our scenario, the enthalpy of neutralization given is -56 kJ/mol for every mole of water formed.

Since we determined that 0.075 moles of KOH will react completely, the same amount of water will be produced. To find the total energy released, multiply the enthalpy change by the moles of water produced:\[\text{Energy released} = (-56 \, \text{kJ/mol}) \times (0.075 \, \text{mol}) = -4.2 \, \text{kJ}\]This shows that -4.2 kJ of energy is released in the reaction. Negative sign indicates the energy is released, making this an exothermic reaction.

Molar Concentration

Molar concentration, often called molarity, is a measure of the concentration of a solute in a solution. It is expressed as moles of solute per liter of solution. In this exercise, concentrations were provided for both HNO_3 and KOH.

To calculate moles from molarity, use the formula:\[\text{Moles} = \text{Molarity} \times \text{Volume} \]Convert volumes from milliliters to liters by dividing by 1000, since molarity is expressed in terms of liters. For HNO_3, it was:\[0.400 \, \text{M} \times 0.200\, \text{L} = 0.080\, \text{mol}\]And for KOH:\[0.500 \, \text{M} \times 0.150 \, \text{L} = 0.075 \, \text{mol}\]Understanding molar concentration helps us approach problems involving solution reactions, enabling us to determine reactants' amount based on provided concentrations and volumes.