Question

In a coffee-cup calorimeter, $150.0\mathrm{mL}$ of $0.50\mathrm{M}$ HCl is added to $50.0\mathrm{mL}$ of $1.00\mathrm{M}\mathrm{NaOH}$ to make $200.0\mathrm{g}$ solution at an initial temperature of ${48.2}^{\circ }\mathrm{C}$. If the enthalpy of neutralization for the reaction between a strong acid and a strong base is $-56\mathrm{kJ}/\mathrm{mol},$ calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is $4.184\mathrm{J}/\mathrm{g}{\cdot }^{\circ }\mathrm{C}$ and assume no heat loss to the surroundings.

Short answer

The final temperature of the calorimeter contents after the reaction between HCl and NaOH is 44.85°C.

Step-by-step solution

Calculate the moles of each reactant

To begin with, we first need to determine the moles of each reactant (HCl and NaOH), using the given molarity and volume. - Moles of HCl: n_HCl = Molarity × Volume = (0.50 mol/L) × (150.0 mL) × $\frac{1L}{1000mL}$ = 0.075 mol - Moles of NaOH: n_NaOH = Molarity × Volume = (1.00 mol/L) × (50.0 mL) × $\frac{1L}{1000mL}$ = 0.050 mol

Determine the limiting reactant

Now we need to identify which reactant is the limiting reactant, since the limiting reactant will determine how much heat is produced during the reaction. Since the moles of NaOH are less than the moles of HCl, the limiting reactant is NaOH.

Calculate the heat released during the reaction

Now we can calculate the heat released during the reaction using the enthalpy of neutralization and the moles of the limiting reactant. Heat released (q) = moles of limiting reactant × enthalpy of neutralization q = 0.050 mol × (-56 kJ/mol) = -2.8 kJ Since the heat is negative, it means that energy is being released during the reaction.

Convert the heat to joules

In order to find the final temperature, we need to have the heat released in joules as the specific heat capacity is given in J/g·°C. q = -2.8 kJ × $\frac{1000J}{1kJ}$ = -2800 J

Calculate the temperature change

Now we can find the temperature change of the solution, which ΔT, by using the heat released (now in joules), the mass of the solution, and the specific heat capacity. ΔT = $\frac{q}{(mass\times specific\mathrm{\_}heat\mathrm{\_}capacity)}$ ΔT = $\frac{-2800J}{(200.0g\times 4.184J/g·°C)}$ = -3.35°C

Calculate the final temperature

Finally, we need to take the initial temperature given and add the calculated temperature change (ΔT) to obtain the final temperature. Final temperature = Initial temperature + ΔT Final temperature = 48.2°C + (-3.35°C) = 44.85°C Thus, the final temperature of the calorimeter contents is 44.85°C.

Key concepts

Enthalpy of Neutralization

The enthalpy of neutralization is a crucial concept in calorimetry, particularly when dealing with strong acids and bases. This refers to the energy change associated with the reaction that occurs when an acid and a base react to form water and a salt. In a calorimetry experiment, like the one described, the enthalpy change helps us understand how much heat is absorbed or released during the reaction at constant pressure. The standard enthalpy of neutralization for strong acids and strong bases is generally around $$-56\text{kJ/mol}$$, which means that this amount of energy is released when one mole of acid reacts with one mole of base. This exothermic reaction releases energy, causing the temperature of the solution to change. Understanding this enthalpy change is essential for calculating final temperatures and understanding energy dynamics in chemical reactions.

Limiting Reactant

The concept of the limiting reactant is pivotal in determining the extent of a chemical reaction. It is the reactant that is consumed first, thus preventing the reaction from continuing because there is no more of it to react. In our scenario, we calculate the moles of each reactant to identify which one is present in smaller amount. - For HCl, we have 0.075 moles. - For NaOH, we have 0.050 moles. Since there are fewer moles of NaOH, it becomes the limiting reactant. This means only 0.050 moles of both HCl and NaOH can be neutralized before the reaction comes to a halt. Understanding which reactant is limiting helps us calculate how much product can be formed and how much energy is released, which is crucial for tasks involving heat exchanges like those in calorimetry.

Specific Heat Capacity

The specific heat capacity of a substance is a measure of how much energy is needed to change the temperature of that substance by one degree Celsius. It is an essential factor in calorimetry.In the given exercise, we assume that the specific heat capacity of the solution is $$4.184\text{J/g}{\cdot }^{\circ }\text{C}$$. This value tells us how much heat energy is needed to elevate the temperature of one gram of the solution by one degree. Using this constant, along with the mass of the solution and the amount of heat exchanged, we are able to calculate the change in temperature. This allows us to find the final temperature of the solution after reaction completion. Understanding the specific heat capacity is crucial because it guarantees accurate calculations in calorimetry experiments.