For the reaction N2(g)+3 H2(g) ⇌2 NH3(g), the equilibrium...

For the reaction $\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g}),$ the equilibrium constant is $K_{p}=36.5$ at $400 \mathrm{K}$. Two separate equilibrium mixtures have the following compositions at $400 \mathrm{K}$ and a total pressure of $1.00 \mathrm{bar}$. Equilibrium mixture A: $0.0424\ \mathrm{mol}\ \mathrm{N}_{2} \quad 0.136\ \mathrm{mol}\ \mathrm{H}_{2} \quad 0.176\ \mathrm{mol}\ \mathrm{NH}_{3}$ Equilibrium mixture B: $0.194\ \mathrm{mol}\ \mathrm{N}_{2} \quad 0.0403\ \mathrm{mol}\ \mathrm{H}_{2} \quad 0.0706\ \mathrm{mol}\ \mathrm{NH}_{3}$ (a) By expressing the partial pressures in the form $P_{i}=x_{i} P,$ where $x_{i}$ is the mole fraction of a particular gas and $P$ is the total pressure, show that the reaction quotient for the reaction can be written as $$ Q_{p}=\frac{x_{\mathrm{NH}_{3}}^{2}}{x_{\mathrm{N}_{2}} x_{\mathrm{H}_{2}}^{3}} \times \frac{1}{P^{2}} $$ [Hint: The mole fraction of $\mathrm{N}_{2}$ is $x_{\mathrm{N}_{2}}=n_{\mathrm{N}_{2}} / n_{\text {tot }}$ Similar expressions hold for $x_{\mathrm{H}_{2}}$ and $x_{\mathrm{NH}_{3}}$. (b) Calculate $Q_{p}$ for each equilibrium mixture to verify that $Q_{p}=K_{p}$ in each case. (c) Suppose that $0.100 \mathrm{mol} \mathrm{N}_{2}$ is added at constant pressure to each mixture. By comparing the values of $Q_{p}$ and $K_{p^{\prime}}$ verify that the addition of $\mathrm{N}_{2}$ causes a net reaction to the right for mixture A but a net reaction to the left for mixture B.

For the reaction $\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g}),$ the
equilibrium constant is $K_{p}=36.5$ at $400 \mathrm{K}$. Two separate equilibrium mixtures have the following compositions at $400 \mathrm{K}$ and a total pressure of $1.00 \mathrm{bar}$. Equilibrium mixture A:
$0.0424\ \mathrm{mol}\ \mathrm{N}_{2} \quad 0.136\ \mathrm{mol}\ \mathrm{H}_{2} \quad 0.176\ \mathrm{mol}\ \mathrm{NH}_{3}$
Equilibrium mixture B:
$0.194\ \mathrm{mol}\ \mathrm{N}_{2} \quad 0.0403\ \mathrm{mol}\ \mathrm{H}_{2} \quad 0.0706\ \mathrm{mol}\ \mathrm{NH}_{3}$
(a) By expressing the partial pressures in the form $P_{i}=x_{i} P,$ where $x_{i}$ is the mole fraction of a particular gas and $P$ is the total pressure, show that the reaction quotient for the reaction can be written as
$$
Q_{p}=\frac{x_{\mathrm{NH}_{3}}^{2}}{x_{\mathrm{N}_{2}} x_{\mathrm{H}_{2}}^{3}} \times \frac{1}{P^{2}}
$$
[Hint: The mole fraction of $\mathrm{N}_{2}$ is $x_{\mathrm{N}_{2}}=n_{\mathrm{N}_{2}} / n_{\text {tot }}$ Similar expressions hold for $x_{\mathrm{H}_{2}}$ and $x_{\mathrm{NH}_{3}}$.
(b) Calculate $Q_{p}$ for each equilibrium mixture to verify that $Q_{p}=K_{p}$ in each case.
(c) Suppose that $0.100 \mathrm{mol} \mathrm{N}_{2}$ is added at constant pressure to each mixture. By comparing the values of
$Q_{p}$ and $K_{p^{\prime}}$ verify that the addition of $\mathrm{N}_{2}$ causes a net reaction to the right for mixture A but a net reaction to the left for mixture B.

Key Concepts

Le Châtelier's Principle and System Perturbation

Le Châtelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions (such as concentration, pressure, or temperature), the system will adjust itself to partially counteract the change and re-establish equilibrium. Adding a reactant or product alters the reaction quotient, and the system will shift in the direction that tends to restore equilibrium based on the comparison between Qp and Kp. This concept is essential for predicting the effects of modifications to the system, such as the addition of reactants in the examples provided.

Reaction Quotient (Qp)

The reaction quotient, Qp, is calculated in the same way as the equilibrium constant but for a system that is not necessarily at equilibrium. By comparing Qp with Kp, one can predict the direction in which the reaction will proceed to achieve equilibrium. If Qp is less than Kp, the reaction will proceed forward (to the right), whereas if Qp is greater than Kp, the reaction will shift in the reverse direction (to the left).

Mole Fraction and Partial Pressure Relationship

In gas-phase equilibria, the partial pressure of each component can be expressed as the product of its mole fraction and the total pressure of the system. This relation is useful for rewriting expressions such as Qp in terms of mole fractions, which simplifies calculations and provides a more standardized form. It connects the macroscopic thermodynamic variables with the microscopic composition of the gas mixture.

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, the concentrations (or partial pressures) of the reactants and products remain constant over time, and this balance is described quantitatively by the equilibrium constant. Understanding chemical equilibrium is crucial for predicting how a system responds to disturbances and for calculating the composition of the system under various conditions.

Equilibrium Constant (Kp)

The equilibrium constant, denoted as Kp for gas-phase reactions, is a dimensionless number that quantifies the ratio of the partial pressures of the products to the reactants at equilibrium, each raised to the power of their stoichiometric coefficients. In the context of this reaction, Kp encapsulates the inherent tendency of the reaction to favor products or reactants at a given temperature, and its value is used to determine the position of equilibrium.

Transcript

00:04 Let's start with our equilibrium here, n2 gas, add 3h2 gas, in equilibrium with 2 nh3 gas.

00:14 We're told that the equilibrium constant is equal to 36 .5 at 400 degrees.

00:23 For part a, we're going to show that the partial pressures can be written as the equation given.

00:38 Qp based on what we have here is the partial pressure of nh3 squared over the partial pressure of n2 times the partial pressure of h2 cubed.

00:50 Now we can rewrite this in terms of mofraction.

00:54 This would be equal to the mole fraction of nh3 times the total partial pressure or the total pressure squared over the mole fraction of n2 times the pressure times the mole fraction of h2 times the pressure cubed if we separate this here this would be equal to the mole fraction of nh3 squared mole fraction of n2 mole fraction of h2 cubed times p squared over p times p cubed.

01:45 And simplifying this, this is equal to, so we can verify that qp here, quotient is equal to a mole fraction of nh3 squared times the mole fraction of n2 times the mole fraction of h2 cubed times one over pressure squared.

02:05 So that verifies the expression that we have.

02:10 Given to us in part a.

02:14 For part b we're asked to verify that equilibrium mixtures a and b both have the quotient equal to the equilibrium constant.

02:32 So before we do that, we're told that the total pressure is 1 bar, which is equal to 0 .987 of an atmosphere.

02:41 Sphere.

02:43 So for equilibrium mixture a, we can see that we have, or we're given, 0 .0424 moles of n2, 0 .136 moles of h2, and 0 .176 moles of nh3.

03:14 Using this here, the most total, by adding everything up would be equal to 0 .3544.

03:25 And using these numbers here, we can calculate the mole fraction of n2, which is equal to 0 .044 moles, divided by 0 .3544.

03:38 This is moles over here, so moles.

03:41 And this would give us 0 .120.

03:44 The mole fraction of h2 is 0 .136.

03:48 136 moles divided by 0 .3544 moles and this works out to 0 .384 and the mole fraction of nh3 is 0 .176 moles over 0 .3544 moles and this will work out to 0 .497.

04:19 Using the equilibrium constant or the quotient expression that we derived in part a, the mole fraction of nh3 squared over the mole fraction of n2 times the mole fraction of h2 cubed times one over the pressure squared.

04:37 Let's substitute our values in and calculate the quotient.

04:40 We'll get 0 .497 squared.

04:44 0 .120 and 0 .384 cubed times 1 over .987 squared.

04:56 We're going to get 35 .6 times 1 .03, which is equal to 36 .5.

05:07 So here we can see that the quotient of 36 .5 is equal to the equilibrium constant given to us in the question at 36 .5.

05:18 So this is equal for equilibrium mixture a.

05:26 For equilibrium mixture b, we'll perform a similar set of calculations.

05:37 We are given that we have 0 .194 moles of n2, 0 .0403 moles of h2, and 0 .0706 moles of nh3.

05:59 The moles total here by adding these all up would be 0 .3049 moles.

06:06 Using this data, we'll calculate the mole fraction of n2, which would be 0 .194 over 0 .3049, and this would yield 0 .636, the mole fraction of h2, 0 .0403 over 0 .39, and this will yield 0 .636, the mole fraction of h2, 0403 over 0 .3049, and this will yield 0 .132, and the mole fraction of nh3, it's 0 .0706 over 0 .3049.

06:49 And this would give us 0 .231.

06:56 A quotient can be calculated as the mole fraction of nh3 squared, mole fraction of n2 times the mole fraction of h2 cubed, times one over the pressure squared.

07:08 Let's substitute our values into this expression here.

07:11 We're going to get 0 .231 squared over, 0 .636 times 0 .132 cubed, 1 over .987 squared.

07:28 This would be equal to 36 .5 times 1 .03.

07:35 And we'll find that the quotient is 37 .6.

07:39 So for equilibrium mixture b, the quotient, which is 37 .6, is not equal to the equilibrium constant of 36 .5.

07:53 It is very close, but it is not equal.

07:58 For c, we're going to add 0 .1 mole of n2 that is added to each, and we're asked to compare the values of the qp and kp, and verify that for the equilibrium mixture a, we're going to shift to the right, but for mixture b, we'll shift to the left.

08:26 So for equilibrium mixture a, we're going to, let's start with our equilibrium, and we have initially 0 .0 -4 -24 moles, 0 .136 moles, and 0 .176 moles...

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