A mixture of NH3( g) and N2 H4(g) is placed in a sealed container at 300 K. The total pressure

A mixture of NH3( g) and N2 H4(g) is placed in a sealed...

A mixture of $\mathrm{NH}_{3}(\mathrm{~g})$ and $\mathrm{N}_{2} \mathrm{H}_{4}(g)$ is placed in a sealed container at $300 \mathrm{~K}$. The total pressure is $0.50 \mathrm{~atm}$. The container is heated to $1200 \mathrm{~K}$ at which time both substances decompose completely according to the equations $2 \mathrm{NH}_{3}(g) \rightarrow \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$; $\mathrm{N}_{2} \mathrm{H}_{4}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) .$ After decomposition is complete the total pressure at $1200 \mathrm{~K}$ is found to be $4.5 \mathrm{~atm}$. Find the percent of $\mathrm{N}_{2} \mathrm{H}_{4}(g)$ in the original mixture. (Assume two significant figures for the temperature.)

A mixture of $\mathrm{NH}_{3}(\mathrm{~g})$ and $\mathrm{N}_{2} \mathrm{H}_{4}(g)$ is placed in a sealed container at $300 \mathrm{~K}$. The total pressure is $0.50 \mathrm{~atm}$. The container is heated to $1200 \mathrm{~K}$ at which time both substances decompose completely according to the equations $2 \mathrm{NH}_{3}(g) \rightarrow \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$;
$\mathrm{N}_{2} \mathrm{H}_{4}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) .$ After decomposition is complete
the total pressure at $1200 \mathrm{~K}$ is found to be $4.5 \mathrm{~atm}$. Find the percent of $\mathrm{N}_{2} \mathrm{H}_{4}(g)$ in the original mixture. (Assume two significant figures for the temperature.)

Key Concepts

Reaction Stoichiometry

Reaction stoichiometry involves using the balanced chemical equations to relate the amounts (in moles) of reactants to those of products. In decomposition reactions, the stoichiometric coefficients indicate the change in mole numbers, which is essential for determining how the composition of a gaseous mixture changes as the reaction proceeds.

Behavior of Gases and the Ideal Gas Law

The ideal gas law (PV = nRT) tells us that under fixed temperature and volume, the pressure of a gas is directly proportional to the number of moles present. This concept is fundamental when analyzing how the total pressure in a sealed container changes due to chemical reactions that change the mole count of gases.

Mole Fraction and Percent Composition

Calculating the percent composition by mole requires determining the ratio of the amount of a specific component to the total amount of all substances present. This concept, often applied in gas mixtures, is used to find the original percentage contribution of a reactant when the system’s total moles change after a reaction.

Transcript

00:01 This question takes a look at some gas properties and gas laws that can be applied to figure some gas -related questions out.

00:09 We're given a mixture of two gases of ammonia and hydrazine, and we're told a couple of facts about them.

00:17 First fact that we're given is that the partial pressure of the ammonia and the partial pressure of the hydrazine together give us a total pressure of 0 .50 atmosphere when my temperature is 300 kelvin.

00:30 Now later on in the question, we're asked to elevate that temperature to 1 ,200 kelvin and we're given some new information.

00:38 The first step we're going to make in this process is just figure out that what the total pressure of this ammonia hydrazine system would be if we elevate the temperature to that 1 ,200 kelvin that we're given in the second part of the question.

00:53 And to do that, we need to use gaila sachs law that says there's a direct relationship between pressure and temperature.

01:00 So here my initial pressure is 0 .50 atmosphere.

01:04 My initial temperature is 300 kelvin.

01:07 And remember, temperatures must be in kelvin in order to utilize gas law relationships.

01:15 My new temperature is going to be 1 ,200 kelvin, and i want to solve here for the new pressure.

01:20 So if i cross multiply and do the math here, ultimately i determine that my new pressure or my total pressure if i were at 1 ,200 kelvin, would be 2 .0 atmosphere.

01:36 Now, in the second part of the problem, i'm told that these two gases decomposed to have a system that has more gaseous particles.

01:46 In fact, i'm told by in the question that when the ammonia decomposes, it produces twice as many particles as it originally had up here.

01:59 And when the hydrazine decompose, decomposes, it produces three times as many gas particles as it had originally.

02:07 And again, remember that the temperature here is that 1 ,200 kelvin.

02:11 And that this new gaseous mixture with those decomposed particles has a total pressure of 4 .5 atmosphere.

02:19 And so because i have twice as many particles of ammonia and three times as many of hydrazine, i know that their partial pressures will increase by that same factor.

02:31 And so i'm just writing this in terms of the original partial pressures of the ammonia and hydrazine.

02:37 And i'm going to approach this now as an algebraic problem, where again, i initially started out by making sure that the original or initial temperature comparison before the decomposition happened would give me a total pressure of the system at the same temperature that i'm going to be comparing the decomposed mixture at.

02:59 So the equation that i'm going to use up here, i'm simply going to replace with a partial pressure of my ammonia and hydrazine at that 2 .0 atmosphere, because that's going to happen at 1 ,200 kelvin or the same temperature at which the second equation or relationship happens when those gas molecules decompose.

03:23 And now what i'm going to do is approach this again as an algebraic problem.

03:27 I'm going to take that equation before the decomposition that i just wrote above with that total pressure at 1 ,200 kelvin.

03:40 So again, it's comparing the same temperature as the above equation here.

03:47 And if i look at this as an algebraic problem and say, okay, i have two variables i want to find out.

03:51 One is the partial pressure of ammonia, the other hydrazine...

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