/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Given that \(K_{c}=61\) for the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Given that \(K_{c}=61\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(500 \mathrm{~K}\), calculate whether more ammonia will tend to form when a mixture of composition \(2.23 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(1.24 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(1.12 \times 10^{-4} \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{NH}_{3}\) is present in a container at \(500 \mathrm{~K}\).

Short Answer

Expert verified
Since the calculated reaction quotient (Q) is greater than the equilibrium constant (Kc), the reaction will shift left, towards reactants, thus more ammonia (NH3) will not form.

Step by step solution

01

- Write the Expression for the Equilibrium Constant (Kc)

For the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3\mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2\mathrm{NH}_{3}(\mathrm{~g})\), the equilibrium constant expression is \(K_{c} = \frac{[\mathrm{NH}_{3}]^{2}}{[\mathrm{N}_{2}][\mathrm{H}_{2}]^{3}}\). This expression relates the concentrations of the reactants and products at equilibrium.
02

- Calculate the Reaction Quotient (Q)

The reaction quotient (Q) is calculated using the initial concentrations of the reactants and products, in the same way as the equilibrium constant: \(Q = \frac{[\mathrm{NH}_{3}]_{\text{initial}}^{2}}{[\mathrm{N}_{2}]_{\text{initial}}[\mathrm{H}_{2}]_{\text{initial}}^{3}}\). Plugging in the given values: \(Q = \frac{(1.12 \times 10^{-4})^{2}}{(2.23 \times 10^{-3})(1.24 \times 10^{-3})^{3}}\).
03

- Calculate Q Value

Calculate the value of Q using the numbers provided: \(Q = \frac{(1.12 \times 10^{-4})^{2}}{(2.23 \times 10^{-3})(1.24 \times 10^{-3})^{3}} = \frac{1.2544 \times 10^{-8}}{4.3215 \times 10^{-12}} = 2.90 \times 10^{3}\).
04

- Compare Q to Kc

Now that we have the Q value, compare it with the equilibrium constant (Kc). The system will shift to reach equilibrium based on whether Q is less than, equal to, or greater than Kc.
05

- Determine the Direction of the Shift

Since the given \(K_{c} = 61\) and the calculated \(Q = 2.90 \times 10^{3}\), and \(Q > K_{c}\), the reaction will shift towards the reactants to reach equilibrium. This means that more ammonia will not tend to form, instead, the concentration of ammonia will decrease.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant (Kc)
Understanding chemical equilibrium is crucial for many fields such as chemistry, engineering, and environmental science. The equilibrium constant (Kc) is a pivotal concept in this regard. It is a numerical value that expresses the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their respective coefficients in the balanced equation. In the context of our example, for the reaction \( \mathrm{N}_2(\mathrm{~g}) + 3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) \), the equilibrium constant expression is \( K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \).

When the value of \( K_c \) is known (in this case, \( K_c = 61 \) at 500 K), it serves as a vital indicator of the extent of the reaction at equilibrium. A larger \( K_c \) suggests that, at equilibrium, the reaction favors products over reactants. Conversely, a smaller \( K_c \) value indicates that the reactants are favored. This helps predict how the system behaves under a set of initial conditions and how the reaction mixture will adjust over time to achieve equilibrium.
Reaction Quotient (Q)
The reaction quotient (Q) is like a snapshot of a reaction that hasn't yet reached equilibrium. It is calculated just like \( K_c \) but uses the current or initial concentrations of the reactants and products instead of the equilibrium concentrations. The formula for the reaction quotient remains the same as that of the equilibrium constant, providing a point of comparison between the current state of the system and the equilibrium state.

For our example reaction, we calculate \( Q \) using the initial concentrations provided: \( Q = \frac{[\mathrm{NH}_3]_{\text{initial}}^{2}}{[\mathrm{N}_2]_{\text{initial}}[\mathrm{H}_2]_{\text{initial}}^{3}} \). By substituting the given concentrations into the equation, we can determine the reaction's move towards equilibrium. If \( Q < K_c \) the reaction will proceed forward, leading to the formation of more product. If \( Q > K_c \) the reaction will shift backward to produce more reactants. Lastly, if \( Q = K_c \) the system is at equilibrium, and no net change will occur.
Le Chatelier's Principle
To further understand how chemical reactions adjust to changes, we turn to Le Chatelier's principle. This principle asserts that if a dynamic equilibrium is disturbed by changing conditions, such as concentration, pressure, or temperature, the system will adjust to counteract the disturbance and restore a new equilibrium.

For instance, if more reactants are added to the system, the principle predicts an increase in the rate of the forward reaction to reduce the reactant concentrations back to equilibrium. If the temperature changes, the reaction could shift in the direction that absorbs heat (in an endothermic reaction) or releases heat (in an exothermic reaction).

In the scenario with our reaction \( \mathrm{N}_2 + 3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 \) and a given \( Q > K_c \), Le Chatelier's principle helps us understand why the reaction shifts towards the reactants when \( Q \) is greater than \( K_c \)—the system is reducing the excess products to return to equilibrium. This elegant principle guides us through the prediction of a reaction's response to various disturbances, ensuring we can effectively control and manipulate chemical processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(400 \mathrm{~K}, K=41\). Find the value of \(\mathrm{K}\) for each of the following reactions at the same temperature. (a) \(2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\) (b) \(\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(2 \mathrm{~N}_{2}(\mathrm{~g})+6 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NH}_{3}(\mathrm{~g})\)

Let \(\alpha\) be the fraction of \(\mathrm{PCl}_{5}\) molecules that have decomposed to \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) in the reaction \(\mathrm{PCl}_{5}(\mathrm{~g})=\) \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) in a constant- volume container, so that the amount of \(\mathrm{PCl}_{5}\) at equilibrium is \(n(1-\alpha)\), where \(n\) is the pressure present initially. Derive an equation for \(K\) in terms of \(\alpha\) and the total pressure \(P\), and solve it for \(\alpha\) in terms of \(P\). Calculate the fraction decomposed at \(556 \mathrm{~K}\), at which temperature \(K=4.96\), and the total pressure is (a) \(0.50\) har; (b) \(1.00\) bar.

A mixture of \(0.0560 \mathrm{~mol} \mathrm{O}_{2}\) and \(0.0200 \mathrm{~mol}\) \(\mathrm{N}_{2} \mathrm{O}\) is placed in a \(1.00-\mathrm{L}\) reaction vessel at \(25^{\circ} \mathrm{C}\). When the reaction \(2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{~g})=4 \mathrm{NO}_{2}(\mathrm{~g})\) is at equilibrium, \(0.0200\) mol \(\mathrm{NO}_{2}\) is present. (a) What are the equilibrium concentrations? (b) What is the value of \(K_{c}\) ?

At \(500^{\circ} \mathrm{C}, K_{c}=0.061\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g})\). If analysis shows that the composition is \(3.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}, 2.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\) and \(0.500 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{NH}_{3}\), is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

(a) Calculate the reaction free energy of \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) when the partial pressures of \(\mathrm{N}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NH}_{3}\) are \(1.0,4.2\), and 63 bar, respectively, and the temperarure is \(400 \mathrm{~K}\). For this reaction, \(K=41\) at \(400 \mathrm{~K}\). (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.