A 0.692-g sample of glucose, C6 H12 O6, was burned in a constant-volume calorimeter. The tempera

step 1 Okay, so the first thing that we can do is we can find delta T. So delta T is the T final, so te

A 0.692-g sample of glucose, C6 H12 O6, was burned in a...

A $0.692-\mathrm{g}$ sample of glucose, $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},$ was burned in a constant-volume calorimeter. The temperature rose from $21.70^{\circ} \mathrm{C}$ to $25.22^{\circ} \mathrm{C}$. The calorimeter contained $575 \mathrm{g}$ of water, and the bomb had a heat capacity of $650 \mathrm{J} / \mathrm{K}$. What is $\Delta U$ per mole of glucose?

Key Concepts

Molar Calculations

This concept involves converting measured energy changes from a given mass of a substance to a per mole basis. It requires understanding the molar mass of the substance and is essential for comparing reaction energetics across different substances and reactions, making it a key step in thermochemical calculations.

Heat Capacity

Heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree. In calorimetry, knowing the heat capacity of components such as water and the calorimeter itself is crucial for calculating the total energy change in the system by relating specific temperature changes to a quantified energy change.

Constant Volume Calorimetry

This concept involves measuring the energy change of a reaction at constant volume, such as in a bomb calorimeter. Since the volume does not change, no work is done by expansion or compression, meaning that all the heat exchanged is directly related to the change in internal energy (?U) of the system.

Internal Energy Change (?U)

In a constant volume process, the heat released or absorbed equals the change in internal energy, ?U. This is because in such systems work due to expansion is zero. ?U is a fundamental thermodynamic property that reflects how the energy inherent within a system changes during a chemical reaction or physical process.

Transcript

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

00:04 So delta t is the t final, so temperature final minus the temperature initial.

00:09 So we know the temperature final is 25 .22 degrees celsius, and we can convert that into kelvin's by adding 273 .15 to it.

00:20 Right, and we do that, we do the same thing to temperature initial, 21 .7 degrees celsius plus 273 .15.

00:28 And we get the answer to be 3 .52 kelvin oh wait 3 .52 kelvin.

00:40 Now the next thing that we can do is calculate the amount of heat absorbed by the water.

00:46 So q water is equal to m c delta t.

00:51 M is 575 grams.

00:56 C is this value so this value is a constant you can look this up in the textbook or online and then delta t is 3 .52.

01:07 Right, kelvin's cancel out, grams cancel out, we're left with joules and that is the units for q.

01:12 So we get 8 .47 times 10 to the third joules.

01:18 Now for q bomb, so that is the amount of heat absorbed by the caliber meter.

01:26 We have c bomb times it by delta t.

01:34 We have 650 joules per kelvin times it by 3 .52.

01:41 Kelvin, 3 .52 kelvin, kelvin cancels out, we're left of joules, so that is our units.

01:50 We get 2 .29 times 10 to the, times 10 to the third joules.

01:59 Now what we need to know is that the amount of energy, q of the reaction is equal to the sum of the q of water plus the q of the bomb, but inverse, right? so, in other words, like, the amount of energy released by the reaction of 0 .692 grams of glucose is equal to the amount of energy absorbed by the water and the bomb.

02:26 Because the amount of energy released by the glucose is only absorbed by the water and the bomb, right? so q of a reaction plus q of water plus q of bomb is equal to zero.

02:41 Because the reaction is releasing energy so it's negative and the water and the bomb is absorbing the energy released by the reaction those are positive and if you add all three of them together they equals zero because the energy is not transferred anywhere else so let's solve for q a reaction since we know what cue of water and q of bomb is so q of water plus q of bomb is is equal to a negative q of reaction.

03:19 Our data was subtracted q of reaction from both sides.

03:22 Now if we divide both sides by negative 1, we get q of reaction is equal to q of water plus q of volume divided by negative 1...

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