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

Short answer

In conclusion, 4.20 kJ of energy will be released when 200.0 mL of 0.400 M HNO3 is mixed with 150.0 mL of 0.500 M KOH.

Step-by-step solution

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the neutralization reaction between HNO3 (a strong acid) and KOH (a strong base). The balanced chemical equation is: \(HNO_{3} + KOH \rightarrow H_{2}O + KNO_{3}\)

Calculate the moles of reactants

Next, we need to find the moles of HNO3 and KOH that are present in the given volumes and concentrations. Moles = Molarity × Volume (in Liters) For HNO3: Moles of HNO3 = 0.400 mol/L × 200.0 mL × (1 L / 1000 mL) = 0.0800 mol For KOH: Moles of KOH = 0.500 mol/L × 150.0 mL × (1 L / 1000 mL) = 0.0750 mol

Determine the limiting reactant

Now, we need to determine the limiting reactant by comparing the mole ratio of reactants to the coefficients in the balanced chemical equation. Mole ratio of HNO3 to KOH = moles of HNO3 / moles of KOH = 0.0800 mol / 0.0750 mol = 1.067 The balanced chemical equation shows a 1:1 mole ratio for the reactants. In this case, KOH has fewer moles than required for reacting with all HNO3 and thus, we identify KOH as the limiting reactant.

Calculate the moles of water produced

Once we have determined the limiting reactant, we can calculate the moles of water produced in the reaction. The balanced chemical equation shows a 1:1 mole ratio between reactants and water, so the moles of water produced will be equal to the moles of the limiting reactant. Moles of water produced = moles of the limiting reactant (KOH) Moles of water produced = 0.0750 mol

Calculate the energy released in the reaction

Finally, we can calculate the energy released in the reaction using the enthalpy of neutralization given, which is -56 kJ/mol of water produced. Energy released = moles of water produced × enthalpy of neutralization per mole of water produced Energy released = 0.0750 mol × (-56 kJ/mol) Energy released = -4.20 kJ In conclusion, 4.20 kJ of energy will be released when 200.0 mL of 0.400 M HNO3 is mixed with 150.0 mL of 0.500 M KOH.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that runs out first and stops the reaction from continuing. It essentially "limits" how much product can be formed. Determining the limiting reactant requires a comparison of the starting moles of each reactant to the stoichiometry of the balanced chemical equation.

In our exercise, we have two reactants: nitric acid, HNO₃, and potassium hydroxide, KOH. We calculated the moles of each from their given concentrations and volumes. This gave us 0.0800 moles of HNO₃ and 0.0750 moles of KOH.

To find the limiting reactant, we compare these moles based on the balanced equation, which shows they react in a 1:1 ratio. Since KOH has fewer moles available (0.0750 moles), it is the limiting reactant. Therefore, the reaction will stop as soon as all the KOH is used up, even if some HNO₃ remains.

Moles of Water Produced

Once we know the limiting reactant, we can determine the moles of water produced. This is because the water is formed according to the same stoichiometric conditions as the reactants in the balanced chemical equation.

In our reaction, each mole of KOH reacts with one mole of HNO₃ to produce one mole of water, H₂O. Since KOH is the limiting reactant and we have 0.0750 moles of it, exactly 0.0750 moles of water are produced.

It’s essential to understand that the stoichiometry of a balanced reaction guides us directly from the limiting reactant to the product quantities. This stoichiometric relationship allows us to confidently know how much product, in this case, water, will result from our specific amounts of reactants.

Enthalpy Calculation

The calculation of energy release or absorption in a chemical reaction can be done using the concept of enthalpy. Enthalpy of neutralization refers to the energy change that occurs when an acid and a base react to form water. It is expressed per mole of water produced.

In this problem, the enthalpy of neutralization is given as \(-56 \mathrm{kJ} / \mathrm{mol}\) of water produced. We found that 0.0750 moles of water are produced, so we can calculate the total energy released by multiplying the moles of water by the enthalpy value:
\[ Energy\ released = 0.0750\ \mathrm{mol} \times (-56\ \mathrm{kJ/mol}) = -4.20\ \mathrm{kJ} \]

The negative sign indicates that the reaction is exothermic, meaning it releases energy to the surroundings. Understanding such calculations is crucial for interpreting energy changes in chemical processes and for practical applications like designing energy-efficient reactions.