Suppose you burned 0.300 g of C(s) in an excess of 𝐎2(g) in a constant-volume calorimeter to giv

Suppose you burned 0.300 g of C(s) in an excess of 𝐎2(g) in...

Suppose you burned 0.300 g of $\mathrm{C}(\mathrm{s})$ in an excess of $\mathbf{O}_{2}(g)$ in a constant-volume calorimeter to give $\mathrm{CO}_{2}(\mathrm{g})$ $$ \mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) $$ The temperature of the calorimeter, which contained 775 g of water, increased from $25.00^{\circ} \mathrm{C}$ to $27.38^{\circ} \mathrm{C} .$ The heat capacity of the bomb is $893 \mathrm{J} / \mathrm{K} .$ Calculate $\Delta U$ per mole of carbon.

Suppose you burned 0.300 g of $\mathrm{C}(\mathrm{s})$ in an excess of $\mathbf{O}_{2}(g)$ in a constant-volume calorimeter to give $\mathrm{CO}_{2}(\mathrm{g})$
$$
\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})
$$
The temperature of the calorimeter, which contained 775 g of water, increased from $25.00^{\circ} \mathrm{C}$ to $27.38^{\circ} \mathrm{C} .$ The heat capacity of the bomb is $893 \mathrm{J} / \mathrm{K} .$ Calculate $\Delta U$ per mole of carbon.

Key Concepts

Molar Energy Conversion

Molar energy conversion involves scaling the measured energy change to a per mole basis. After finding the total energy change from the reaction, one uses the amount of substance (in moles) that underwent the reaction to determine the energy change associated with one mole of a reactant, providing a standardized measure of reaction energetics.

Calorimeter Heat Capacity

Calorimeter heat capacity is a measure of how much heat energy is required to raise the temperature of the calorimeter by one degree. It is an essential parameter in calorimetry because it allows the conversion of observed temperature changes into an amount of heat absorbed or released during the reaction.

Heat Transfer

Heat transfer involves the movement of thermal energy from one substance or system to another and is governed by the conservation of energy. In calorimetric experiments, the heat produced by a chemical reaction is absorbed by the surrounding materials, such as water and the calorimeter itself, leading to a measurable increase in temperature.

Constant Volume Calorimetry

This concept refers to the measurement of energy changes in a reaction that occurs in a closed system where the volume remains constant. In such a setup, any energy change is manifested as a change in the internal energy of the system, allowing one to directly relate the heat released or absorbed during the reaction to the change in internal energy (?U).

Transcript

00:01 Okay, so the first thing that we should do is we can calculate the change in temperature.

00:07 So delta t is equal to t final, so the temperature final minus the temperature initial.

00:12 Right, so we know the temperature final is 27 .38 degrees celsius.

00:18 We need to convert that into kelvin's.

00:21 And the conversion between celsius and kelvin is you just got to add 273 to the celsius.

00:26 273 .15 to the celsius.

00:28 So this is the final in kelvin.

00:32 Now let's subtract the final, the initial in kelvin.

00:34 So we have 25 degrees celsius plus 273 .15.

00:40 And we get the change in temperature to be equal to 2 .38 kelvin.

00:50 Now the next thing that we can do is we can calculate the amount of heat absorbed by the water and then calculate the amount of heat absorbed by the bomb.

01:00 Right.

01:00 So the amount of heat absorbed by the water is q is equal to mc.

01:03 Delta t, right? we know the m is 775 grams.

01:09 That's how much water we have.

01:12 We know c is 4 .184 joules per gram times kelvin.

01:18 This was not given to us in the question, but it's a constant so you can look it up in the textbook.

01:26 And delta t we calculated 2 .38 kelvin, right, kelvin cancels, grams cancels, and we get jewels and that is the unit for q.

01:35 We get the answer to be 7 .72 times 10 to the 3rd joules.

01:46 Now the next thing that we can do is calculate the q bomb, right? so q bomb is equal to the specific heat of the bomb times by delta t.

02:02 So we get 893.

02:06 Let me rewrite this.

02:10 893 joules per kelvin...