Question

When \(35.0 \mathrm{~mL}\) of \(1.43 \mathrm{M} \mathrm{NaOH}\) at \(22.0^{\circ} \mathrm{C}\) is neutralized by \(35.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) also at \(22.0^{\circ} \mathrm{C}\) in a coffee-cup calorimeter, the temperature of the final solution rises to \(31.29^{\circ} \mathrm{C}\). Assume that the specific heat of all solutions is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), that the density of all solutions is \(1.00 \mathrm{~g} / \mathrm{mL}\), and that volumes are additive. (a) Calculate \(q\) for the reaction. (b) Calculate \(q\) for the neutralization of one mole of \(\mathrm{NaOH}\).

Short answer

Question: Calculate (a) the heat of the reaction and (b) the heat of neutralization of one mole of NaOH for the given reaction. Answer: (a) The heat of the reaction (q) is 2725 J. (b) The heat of neutralization of one mole of NaOH is 54,446 J/mol.

Step-by-step solution

Calculate the total volume and mass of the solution

According to the problem, volumes are additive. We have 35.0 mL of NaOH solution and 35.0 mL of HCl solution. Therefore, the total volume is: Total volume = Volume of NaOH + Volume of HCl = 35.0 mL + 35.0 mL = 70.0 mL Since the density of all solutions is 1.00 g/mL, we can calculate the mass of the solution: Total mass = Total volume × Density = 70.0 mL × 1.00 g/mL = 70.0 g

Calculate the change in temperature

We are given the initial temperature (22.0°C) and the final temperature (31.29°C). To find the change in temperature, we subtract the initial temperature from the final temperature: ∆T = T_final - T_initial = 31.29°C - 22.0°C = 9.29°C

Calculate the heat of the reaction (q)

Using the formula: q = mc∆T, we can calculate the heat of the reaction: q = (70.0 g)(4.18 J/g°C)(9.29°C) = 2724.974 J We can round this to: q = 2725 J

Calculate the moles of NaOH

We are given the volume (35.0 mL) and molarity (1.43 M) of the NaOH solution. To find the moles of NaOH, we can use the formula: moles = molarity × volume, but we need to convert volume to liters first: Volume = 35.0 mL × (1 L / 1000 mL) = 0.035 L Now we can calculate the moles of NaOH: Moles of NaOH = 1.43 mol/L × 0.035 L = 0.05005 mol

Calculate the heat of neutralization of one mole of NaOH

Finally, with the heat of the reaction (q) and the moles of NaOH, we can calculate the heat of neutralization of one mole of NaOH: Heat of neutralization = q / moles of NaOH = 2725 J / 0.05005 mol = 54445.55 J/mol We can round this to: Heat of neutralization = 54,446 J/mol In conclusion, (a) The heat of the reaction (q) is 2725 J. (b) The heat of neutralization of one mole of NaOH is 54,446 J/mol.

Key concepts

Calorimetry

Calorimetry is a method used to measure the amount of heat transferred in a chemical reaction, such as the neutralization process in our exercise. It's like a scientific way to measure how hot or cold a chemical reaction can make the surroundings. Imagine stirring a cup of tea with a spoon; you can feel the temperature change. In calorimetry, we use special tools called calorimeters to measure this change in heat more precisely. The instrument helps us understand whether the reaction absorbs heat (feeling cooler, like sucking on a mint) or releases heat, warming up like a cozy blanket.

In our textbook example, a coffee-cup calorimeter is used, which is a simple kind of calorimeter. We could think of it as a thermos that measures how much the temperature goes up or down when an acid and a base are mixed together. It's based on the principle that the heat given off or absorbed by the reaction (the solution in the cup) will change the temperature of the water in the cup. By knowing how the temperature changes, the specific heat (that's how much energy it takes to change the temperature of a certain amount of substance), and the mass, we can calculate the heat of the reaction using the formula q = mcΔT.

Enthalpy Change

Enthalpy change represents the heat change at constant pressure and is denoted as ΔH. It's like a scorecard telling us if a reaction is putting heat out or sucking it in. This is crucial because it helps us understand which reactions can keep us warm or cool, similar to when you burn wood in a fireplace (releases heat) or when you dissolve ammonium nitrate in water to make a cold pack (absorbs heat).

In the context of our problem, the enthalpy change of the neutralization reaction is what we're measuring. For the neutralization process, where an acid and a base cancel each other out (it's like when enemies in movies decide to shake hands), the enthalpy change is often exothermic, meaning it releases heat (similar to a pleasant campfire). In the example we're dealing with, the heat of neutralization can be thought of as how much energy is let out when one mole of sodium hydroxide (NaOH) reacts with an acid to form water and a salt.

Molarity and Solution Stoichiometry

When you bake a cake, you need to measure ingredients carefully. You wouldn't want too much or too little of anything! Molarity and solution stoichiometry are a bit like that, but for chemical reactions. Molarity tells us how many moles of stuff (like how many packets of cake mix) are in a certain volume of liquid (like how much water to stir in). It's typically written as moles per liter (mol/L) and noted with the letter 'M'.

Solution stoichiometry, meanwhile, is the part that's like following the recipe. It tells us how the amounts of different ingredients (reactants) combine and what amounts of the products they'll make. In our exercise, we calculate the molarity to find out how many moles of sodium hydroxide we have, and use stoichiometry to understand how it'll react with the hydrochloric acid, and finally, to determine how much heat (q) this reaction produces per mole of reactants. This helps us make predictions, like how a particular amount of baking soda will make a cake rise.

Endothermic and Exothermic Reactions

Reactions are either party starters or party poopers. The party starters, or exothermic reactions, release heat, lighting up the surroundings like a firework show. The party poopers, or endothermic reactions, are the quiet ones, absorbing heat, leaving the surrounding a bit colder, like a snowball in your hand.

In our textbook example, the increase in temperature from the reaction indicates that it is exothermic. The combination of NaOH and HCl doesn't just create water and a salt; it also lets out energy in the form of heat, like a crowd cheering and warming up the room. The measurement of this heat release helps us understand the energy dynamics of the chemical process. The feeling of warmth when neutralizing an acid with a base is due to the reaction giving off energy, making the heat of neutralization a helpful concept in both teaching us about chemical reactions and keeping us cozy!