When 10.00 g of phosphorus is burned in O2(g) to form P4 O10(s), enough heat is generated to rai

step 1 So I'm going to be burning phosphorus in air. Let's go ahead and write out the chemical equation

When 10.00 g of phosphorus is burned in O2(g) to form P4...

Key Concepts

Calorimetry

Calorimetry is the experimental measurement of the heat exchanged by a system during a chemical or physical process. In the context of this reaction, the burning of phosphorus generates heat that is absorbed by a known mass of water. By measuring the temperature change of the water and knowing its specific heat capacity, one can calculate the amount of heat transfer associated with the reaction.

Specific Heat Capacity and Energy Calculation

The specific heat capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. This concept is used to calculate the heat absorbed by the water by multiplying the mass of the water, its specific heat capacity, and the change in temperature. This total energy calculation is essential for linking the calorimetric data to the thermodynamic properties of the chemical reaction.

Enthalpy of Formation

The enthalpy of formation refers to the heat change associated with the formation of one mole of a compound from its elements in their standard states. In this scenario, the heat generated by burning phosphorus in oxygen is used to determine how much heat is released per mole of P4O10 formed. This concept is critical for understanding the thermodynamic stability and the energy requirements or releases associated with compound formation.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. In this problem, stoichiometry is used to convert the mass of phosphorus burned into moles, and subsequently, relate the measured heat release to the number of moles of P4O10 produced. This conversion is necessary to express the enthalpy change on a per-mole basis, which is a standard way of reporting thermodynamic data.

Transcript

00:01 So i'm going to be burning phosphorus in air.

00:05 Let's go ahead and write out the chemical equation for that.

00:09 So phosphorus, solid, plus oxygen gas.

00:18 It's going to give me p410, and that's going to be a solid.

00:30 Lovely.

00:30 So let's go ahead and balance this.

00:32 I know i'm going to need four phosphorus.

00:36 Here, and i'll need five pairs of oxygen atoms to give me 10 overall.

00:44 Great.

00:45 So i'm going to be using the heat that's given off by this reaction to warm up some water.

00:51 And so we can say that the heat of the reaction, or the heat released in the reaction, is then equal to the amount of heat that's absorbed by the water.

01:03 Okay, so let's go ahead and do the water portion.

01:06 So, q water.

01:09 Is going to be equal to, well, the mass of the water, which we're told is 2 ,950 grams, times the specific heat of water.

01:21 So 4 .18 joules per gram degrees celsius times the change in temperature, which is 38 degrees celsius minus 18 degrees celsius.

01:37 All right.

01:39 That gives us 2 .466 times 10 to the fifth joules.

01:49 That's the same thing as 2 .466 times 10 to the second kilojoules.

02:00 Okay, that is the heat absorbed by the water.

02:05 We can do the heat absorbed, or sorry, release.

02:09 By the reaction then is going to be the heat of the reaction times the number of moles involved now heat of the reaction let's go back to the original equation here remember heat of reaction is equal to enthalpy of formation of the products minus enthalpy of formation of the reactants but notice that both the reactants here are in their natural state.

02:49 So they're going to have enthalpyes of formation of zero.

02:52 So this goes to zero...

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