Insoluble PbBr2(s) precipitates when solutions of Pb(NO3)2(aq) and NaBr(aq) are mixed. Pb(NO3)2(

Insoluble PbBr2(s) precipitates when solutions of...

Insoluble $\mathrm{PbBr}_{2}(\mathrm{s})$ precipitates when solutions of $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})$ and $\mathrm{NaBr}(\mathrm{aq})$ are mixed. $$\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaBr}(\mathrm{aq}) \rightarrow \mathrm{PbBr}_{2}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})$$ To measure the enthalpy change, $200 .$ mL of $0.75 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})$ and $200 . \mathrm{mL}$ of $1.5 \mathrm{M}$ NaBr(aq) are mixed in a coffee-cup calorimeter. The temperature of the mixture rises by $2.44^{\circ} \mathrm{C}$ Calculate the enthalpy change for the precipitation of $\mathrm{PbBr}_{2}(\mathrm{s}),$ in $\mathrm{kJ} / \mathrm{mol}$. (Assume the density of the solution is $1.0 \mathrm{g} / \mathrm{mI}_{7}$ and its specific heat capacity is $4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .)$

Insoluble $\mathrm{PbBr}_{2}(\mathrm{s})$ precipitates when solutions of $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})$ and $\mathrm{NaBr}(\mathrm{aq})$ are mixed.
$$\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaBr}(\mathrm{aq}) \rightarrow \mathrm{PbBr}_{2}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})$$
To measure the enthalpy change, $200 .$ mL of $0.75 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})$ and $200 . \mathrm{mL}$ of $1.5 \mathrm{M}$
NaBr(aq) are mixed in a coffee-cup calorimeter. The temperature of the mixture rises by $2.44^{\circ} \mathrm{C}$ Calculate the enthalpy change for the precipitation of $\mathrm{PbBr}_{2}(\mathrm{s}),$ in $\mathrm{kJ} / \mathrm{mol}$. (Assume the density of the solution is $1.0 \mathrm{g} / \mathrm{mI}_{7}$ and its specific heat capacity is $4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .)$

Key Concepts

Precipitation Reactions

Precipitation reactions occur when two soluble salts in solution react to form an insoluble solid, or precipitate. In this reaction, mixing lead nitrate and sodium bromide results in the formation of insoluble lead bromide. Understanding precipitation is important in thermochemistry because the formation of the solid may involve unique heat exchanges that need to be measured and calculated per mole of the precipitate.

Enthalpy Change

Enthalpy change (?H) is a measure of the heat absorbed or released during a chemical reaction at constant pressure. In this problem, the calculated energy change from the calorimeter experiment is converted into an enthalpy change per mole of the precipitate formed. It quantifies the overall energy exchange associated with the precipitation process.

Calorimetry

Calorimetry is the experimental technique used to measure the amount of heat released or absorbed during a chemical reaction. In this scenario, a coffee-cup calorimeter is employed to measure the temperature change when solutions are mixed, allowing the calculation of the heat change (q) based on the mass of the solution, its specific heat capacity, and the observed temperature change.

Stoichiometry

Stoichiometry involves using the balanced chemical equation to determine the quantitative relationships between reactants and products. In the context of this problem, stoichiometry is used to find the limiting reactant and to relate the amount of precipitate formed to the energy measured by the calorimeter, ensuring that the calculated enthalpy change is expressed per mole of product formed.

Specific Heat Capacity

Specific heat capacity is a property of a substance that indicates the amount of energy required to raise the temperature of one gram of the substance by one degree Celsius (or Kelvin). This value is essential in calorimetry calculations, as it allows the conversion of temperature change into energy (q = m × c × ?T), thus linking the thermal measurement to the enthalpy change of the reaction.

Transcript

00:01 So for this question, we are trying to calculate the change in enthalpy for this precipitation reaction.

00:11 So the equation that we're going to use is q, which is going to represent our heat or change in enthalpy, is equal to the mass times the specific heat times the change in temperature.

00:24 So this is going to tell us the enthalpy change for the reaction, or one mole of the reaction.

00:40 So q is what we're trying to find.

00:45 The mass, since we have two solutions, if the mass is going to be the mass of both those solutions added together, we're already given our specific heat and we're already given our change in temperature.

01:02 So before we get started, let's use the density to turn these milliliters into grams.

01:06 So we take it times the density, one gram per milliliter, it's going to be 200 grams.

01:13 Same for this one, we take it times the density, so it's going to be 200 grams.

01:20 So our total mass, 200 grams plus 200 grams, is going to be 400 grams.

01:27 So let's plug that into our equation.

01:29 Our q is going to be equal to the mass, 400 grams, times the specific heat of the solution, 4 .2 joules per grams times celsius, times the change in temperature, which was 2 .44 degrees celsius.

01:46 Type that into your calculator, you should get 4099 joules.

01:56 So this is for, so now we need to calculate the change in enthalpy for when the precipitation of, or the solid of the lead bromide is.

02:18 So before we do that, it asks us to do the enthalpy change in kilojoules per mole.

02:24 So first let's convert joules to kilojoules.

02:27 So for every one kilojoule, we have a thousand joules.

02:31 Joules cancel out, we're gonna have 4 .099 kilojoules, and that's for one mole of the reaction.

02:50 We gotta figure out what it's gonna be for the specific amount of lead bromide that is precipitated.

03:01 So in order to do that, we need to figure out how much lead bromide is precipitated.

03:11 So the change in enthalpy for the reaction is going to be the heat change divided by the number of moles of our precipitate.

03:24 So the number of moles of lead bromide.

03:29 So we've already figured out our q.

03:34 Now we need to figure out how many moles of the precipitate formed...

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