Sulfur (2.56 g) was burned in a constant-volume calorimeter...

Sulfur $(2.56 \mathrm{g})$ was burned in a constant-volume calorimeter with excess $\mathbf{O}_{2}(\mathrm{g}) .$ The temperature increased from $21.25^{\circ} \mathrm{C}$ to $26.72^{\circ} \mathrm{C}$. The bomb has a heat capacity of $923 \mathrm{J} / \mathrm{K},$ and the calorimeter contained $815 \mathrm{g}$ of water. Calculate $\Delta U$ per mole of $\mathrm{SO}_{2}$ formed for the reaction $$ s_{s}(s)+8 \mathbf{O}_{2}(g) \rightarrow 8 \mathbf{S O}_{2}(\mathbf{g}) $$ (IMAGE CANNOT COPY)

Sulfur $(2.56 \mathrm{g})$ was burned in a constant-volume calorimeter with excess $\mathbf{O}_{2}(\mathrm{g}) .$ The temperature increased from $21.25^{\circ} \mathrm{C}$ to $26.72^{\circ} \mathrm{C}$. The bomb has a heat capacity of $923 \mathrm{J} / \mathrm{K},$ and the calorimeter contained $815 \mathrm{g}$ of water. Calculate $\Delta U$ per mole of $\mathrm{SO}_{2}$ formed for the reaction
$$
s_{s}(s)+8 \mathbf{O}_{2}(g) \rightarrow 8 \mathbf{S O}_{2}(\mathbf{g})
$$
(IMAGE CANNOT COPY)

Key Concepts

Internal Energy Change (?U)

In constant-volume processes, the heat measured corresponds directly to the change in the internal energy of the system. This concept is central to thermodynamics as it connects the measured temperature change and heat capacity of the calorimeter to the overall energy change from the chemical reaction, allowing calculation of ?U per unit of reactant or product.

Stoichiometry in Thermochemistry

Stoichiometry ensures that the energy change computed from the calorimetric data can be related to the number of moles of a particular substance involved in the reaction. It is essential for converting the measured total energy change into a per-mole value for a reactant or product, thereby allowing meaningful comparisons and further analysis in chemical thermodynamics.

Constant-Volume Calorimetry

This technique involves carrying out chemical reactions in a rigid, non-volume-changing container so that any heat change is directly related to the internal energy change of the reaction. It is particularly useful for reactions where work done by expansion is negligible and the energy change measured corresponds to changes in the chemical energy of the system.

Bomb Calorimeter and Heat Capacity

Bomb calorimetry is a type of constant-volume calorimetry where reactions, often combustion processes, are carried out in a sealed, high-pressure container. The total heat capacity of the calorimeter, which includes contributions from both the calorimeter itself and any substances like water, is critical in calculating the amount of heat absorbed or released during the reaction, thereby determining the reaction’s energy change.

Transcript

00:01 Okay, so the first thing that we can do is we can calculate delta t.

00:04 Right, so delta t is equal to the temperature of the final minus the temperature of the initial, right? so we have final temperature is 26 .72 degrees celsius.

00:17 And we know the conversion is plus 273 .15 in order to convert celsius to kelvin, right? so we do the same thing for the t initial.

00:28 Oops 21 .25 degrees celsius plus 273 .15.

00:36 Right, and we get delta t to be equal to 5 .47 kelvin.

00:43 Now the next thing that we can do is we can calculate how much heat is transferred to the water, and then we can calculate the amount of heat transferred to the bomb.

00:53 Right, so heat transfer to the water is equal to mc delta t, mc delta t we know m is 815 grams of water you know c is 4 .184 joules per grams times kelvin this you can find in the textbook it's a constant for water and then delta t we found right 5 .47 kelvin and we get the answer to be 1 .87 times 10 to the 4 joules so that is the amount of heat transferred to water.

01:33 Now the next thing we can do is calculate the amount of heat transfer to calculate the amount of heat transfer to the bottom, right? so we have the specific heat, 923 joules per kelvin, and we have the change in temperature is 5 .47 kelvin.

01:54 And we get the answer to be 5 .05 times 10 to the 3rd joules.

02:01 So we know that the amount of, of heat released by the reaction plus the amount of heat gained by the bomb and the water is equal to zero, right? so just means that q of reaction plus q of water plus q of the bomb is equal to zero because the amount of heat lost by the reaction is gained by the water and bomb only.

02:36 So this means that the amount of heat loss by the reaction must be negative.

02:42 So let's solve for q of reaction since we know what the q of water and the q of bomb is.

02:48 So we get q of water plus q of bomb is equal to negative q of reaction.

03:01 Negative q of reaction.

03:05 So all i did was subtract the q of reaction from both sides of the equation.

03:12 And now let's divide both sides by negative 1 since we're trying to solve for q of.

03:17 Reaction right so divide both sides by negative 1 and we get q of reaction is equal to q of water plus q of the bomb divided by negative 1 now let's plug in the values so we have q of water is 1 .87 times 10 to the 4 joules plus 5 .05 times 10 to the 3rd jules divided by negative 1 and we should get the answer to be negative 23 .8 times 10 to the third joules...

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