Question Video: Calculating 𝐾_𝑝 at Equilibrium for a Mixture of Nitrogen, Hydrogen, and Ammonia | Nagwa Question Video: Calculating 𝐾_𝑝 at Equilibrium for a Mixture of Nitrogen, Hydrogen, and Ammonia | Nagwa

Question Video: Calculating ๐พ_๐‘ at Equilibrium for a Mixture of Nitrogen, Hydrogen, and Ammonia Chemistry • Third Year of Secondary School

6.00 mol of Nโ‚‚ gas and 20.00 mol of Hโ‚‚ were allowed to react at 650 K and 50 atm of pressure. At equilibrium, 4.00 mol of Nโ‚‚ gas had been converted into ammonia according to the following equation: Nโ‚‚ (g) + 3 Hโ‚‚ (g) โ‡Œ 2 NHโ‚ƒ (g). Calculate ๐พ_๐‘ for this equilibrium, giving your answer to two decimal places in scientific notation.

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Video Transcript

6.00 moles of N2 gas and 20.00 moles of H2 were allowed to react at 650 kelvin and 50 atmospheres of pressure. At equilibrium, 4.00 moles of N2 gas had been converted into ammonia according to the following equation: N2 gas plus 3H2 gas reacting reversibly to give 2NH3 gas. Calculate ๐พ ๐‘ for this equilibrium, giving your answer to two decimal places in scientific notation.

We are asked to calculate ๐พ ๐‘. ๐พ ๐‘ is a special equilibrium constant, which we can use when the substances in a reaction are gases, when the reaction is reversible and at equilibrium, and this can only take place in a closed system. ๐พ ๐‘ is the ratio of the concentrations of the products and reactants but expressed as partial pressures.

In this problem, ๐พ ๐‘ is equal to the partial pressure of the product, ammonia, NH3, raised to the power of two, since ammoniaโ€™s stoichiometric coefficient is two. The denominator will be the partial pressure of the reactant, N2, nitrogen gas, raised to the power of one, since nitrogenโ€™s stoichiometric coefficient is one, multiplied by the partial pressure of the other reactant, H2, which is hydrogen gas, raised to the power of three, since this substance has a stoichiometric coefficient of three.

We have now formulated the ๐พ ๐‘ expression for this reversible reaction. It is not necessary to show powers of one, and so to simplify, letโ€™s remove it. If we choose, we can simplify further by removing these parentheses here. To calculate ๐พ ๐‘, all we need are the partial pressures of ammonia, nitrogen, and hydrogen. But we are not given these values. So we have to use data that we have been given to first calculate these three partial pressures and then calculate ๐พ p.

Letโ€™s clear some space to do the calculation. We can draw a handy table to see what data we have and what is missing. We are told that six moles of nitrogen and 20 moles of hydrogen react. We can assume that initially there is no ammonia. We are also told that at equilibrium four moles of nitrogen have been converted or changed into ammonia.

Because we know the starting number of moles of the reactants and the moles of one of the reactants that has been converted or changed, we can use an ICE table, where I is the initial or starting number of moles, C the changed or converted moles, and E the equilibrium number of moles for each species. We can fill in these initial values. Changed or converted moles are the number of moles of each substance that reacted or were produced. We know that 4.00 moles of nitrogen are converted to ammonia. But how many moles of hydrogen reacted and how many moles of ammonia were produced overall? We need to use the mole ratio one as to three as to two to calculate these values.

Since the ratio of nitrogen to ammonia is one as to two, then if 4.00 moles of nitrogen reacted or converted or changed, then 8.00 moles of ammonia must have been produced. In a similar way, if the moles of nitrogen to hydrogen is one as to three, then the moles of hydrogen that changed must have been three times that of the moles of nitrogen which were changed, which gives 12.00 moles of hydrogen that reacted.

The third line on the ICE table, equilibrium moles, is the number of moles of each species present at equilibrium. Initially, there were 6.00 moles of nitrogen. 4.00 moles of this were changed or reacted. And so the difference, 2.00 moles, must be the moles of nitrogen present at equilibrium. And this is the number of moles of nitrogen that did not react. Doing the same calculation for the other reactant, we get 8.00 moles of hydrogen present at equilibrium or unreacted. Initially, there was no ammonia in the system. At the end of the reaction, 8.00 moles of ammonia had been produced. This means at equilibrium there were 8.00 moles of ammonia present. Can you see that for the product we are doing an addition and not a subtraction?

We are now going to use the equilibrium number of moles of each species to calculate the partial pressure for each gas. But this is a two-step process. First, we need to calculate the mole fraction of each substance and, from this, the partial pressure of each substance. Then, we can calculate ๐พ ๐‘. To calculate the mole fraction of a substance, letโ€™s call it ๐ด, we need to know the number of moles of that substance at equilibrium divided by the total number of moles of all the substances.

If we take the sum of these three values, we will calculate the total number of moles of particles in the system at equilibrium. Performing this sum, we get a total of 18.00 moles. Therefore, the denominator value for the mole fraction calculation for each substance is 18.00 moles. For nitrogen, the numerator value is 2.00 moles, and for hydrogen and ammonia, 8.00 moles. Solving for each, we get 0.111 recurring for nitrogen, 0.444 recurring for hydrogen, and 0.444 recurring for ammonia. Letโ€™s keep too three significant figures for simplicity.

Note that mole units will cancel, and mole fraction is a unitless or dimensionless quantity. These values are the fraction of the total number of moles present at equilibrium. There is about 11.1 percent of nitrogen present at equilibrium and about 44.4 percent hydrogen and 44.4 percent ammonia.

Now we can use the equation the partial pressure of a substance, letโ€™s call it ๐ด, is equal to the mole fraction of that substance multiplied by the total pressure. We know the mole fraction of each substance at equilibrium. And we are told the total pressure is 50 atmospheres. So nitrogenโ€™s partial pressure is equal to 0.111, the mole fraction, multiplied by 50 atmospheres, giving us 5.55 atmospheres, the partial pressure of nitrogen. This value means that of the total pressure of 50 atmospheres, 5.55 atmospheres comes from nitrogen. We can do a similar calculation for hydrogen and ammonia. And we get a partial pressure for each of 22.2 atmospheres.

Finally, we are ready to calculate ๐พ ๐‘. We can substitute the partial pressures for each substance into the ๐พ ๐‘ expression. And we get a ๐พ ๐‘ value for this reaction in scientific notation to two decimal places, which is what we were asked, of 8.12 times 10 to the negative three. We have not included units because ๐พ ๐‘ is normally a dimensionless or unitless quantity.

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