Skip to main content
Ch.15 - Chemical Equilibrium
Brown - Chemistry: The Central Science 15th Edition
Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970๋‹น์‹ ์ด ์‚ฌ์šฉํ•˜๋Š” ๊ฒŒ ์•„๋‹ˆ๋ผ์š”?๊ต๊ณผ์„œ ๋ณ€๊ฒฝ
๋ชจ๋“  ๊ต๊ณผ์„œBrown 15th EditionCh.15 - Chemical Equilibrium๋ฌธ์ œ 28
15์žฅ, ๋ฌธ์ œ 28

Consider the equilibrium N2(๐‘”) + O2(๐‘”) + Br2(๐‘”) โ‡Œ 2 NOBr(๐‘”) Calculate the equilibrium constant ๐พ๐‘ for this reaction, given the following information at 298 K:
2 NO(๐‘”) + Br2(๐‘”) โ‡Œ 2 NOBr(๐‘”) ๐พ๐‘ = 2.02
NO(๐‘”) โ‡Œ N2(๐‘”) + O2(๐‘”) ๐พ๐‘ = 2.1ร—1030

๊ฒ€์ฆ๋œ ๋‹จ๊ณ„๋ณ„ ์•ˆ๋‚ด
1
Identify the given reactions and their equilibrium constants. The first reaction is 2 NO(g) + Br2(g) โ‡Œ 2 NOBr(g) with Kc = 2.02. The second reaction is NO(g) โ‡Œ N2(g) + O2(g) with Kc = 2.1ร—10^30.
Write the target reaction in terms of the given reactions. The target reaction is N2(g) + O2(g) + Br2(g) โ‡Œ 2 NOBr(g). Notice that this can be achieved by reversing the second reaction and then adding it to the first reaction.
Calculate the equilibrium constant for the reversed second reaction. For a reaction that is reversed, the equilibrium constant is the reciprocal of the original. Thus, Kc for the reversed reaction N2(g) + O2(g) โ‡Œ NO(g) is 1/(2.1ร—10^30).
Combine the equilibrium constants of the adjusted reactions to find the equilibrium constant for the target reaction. When reactions are added, their equilibrium constants are multiplied. Therefore, multiply the Kc of the reversed second reaction with the Kc of the first reaction.
Convert the combined Kc to Kp using the relation Kp = Kc(RT)^(ฮ”n), where ฮ”n is the change in moles of gas between reactants and products, R is the gas constant, and T is the temperature in Kelvin. Calculate ฮ”n for the target reaction and use it to find Kp.

๋น„์Šทํ•œ ๋ฌธ์ œ์— ๋Œ€ํ•œ ๊ฒ€์ฆ๋œ ์˜์ƒ ๋‹ต๋ณ€:

์ด ์˜์ƒ ํ•ด๋ฒ•์€ ์œ„ ๋ฌธ์ œ์— ๋„์›€์ด ๋œ๋‹ค๊ณ  ํŠœํ„ฐ๋“ค์ด ์ถ”์ฒœํ•œ ๊ฒƒ์ž…๋‹ˆ๋‹ค.
์˜์ƒ ๊ธธ์ด:
6m

์ฃผ์š” ๊ฐœ๋…

์งˆ๋ฌธ์— ์˜ฌ๋ฐ”๋ฅด๊ฒŒ ๋‹ตํ•˜๊ธฐ ์œ„ํ•ด ๋ฐ˜๋“œ์‹œ ์ดํ•ดํ•ด์•ผ ํ•˜๋Š” ํ•ต์‹ฌ ๊ฐœ๋…๋“ค์€ ๋‹ค์Œ๊ณผ ๊ฐ™์Šต๋‹ˆ๋‹ค.

Equilibrium Constant (K)

The equilibrium constant (K) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. It is specific to a particular reaction at a certain temperature. For reactions involving gases, the equilibrium constant can be expressed in terms of partial pressures (Kp) or concentrations (Kc), depending on the state of the reactants and products.
์ถ”์ฒœ ์˜์ƒ:
๊ฐ€์ด๋“œ ์ฝ”์Šค
01:14
Equilibrium Constant K

Relationship Between Kp and Kc

The relationship between Kp and Kc is defined by the equation Kp = Kc(RT)^(ฮ”n), where R is the ideal gas constant, T is the temperature in Kelvin, and ฮ”n is the change in the number of moles of gas between products and reactants. This relationship allows for the conversion of equilibrium constants from concentration-based to pressure-based values, which is essential when dealing with gaseous reactions.
์ถ”์ฒœ ์˜์ƒ:
๊ฐ€์ด๋“œ ์ฝ”์Šค
02:18
Kp vs Kc Relationship

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust itself to counteract the change and restore a new equilibrium. This principle is crucial for understanding how changes in concentration, pressure, or temperature affect the position of equilibrium and the values of Kp and Kc in a chemical reaction.
์ถ”์ฒœ ์˜์ƒ:
๊ฐ€์ด๋“œ ์ฝ”์Šค
07:32
Le Chatelier's Principle
๊ด€๋ จ ์‹ค์ฒœ
๊ต๊ณผ์„œ ์งˆ๋ฌธ

At 1000 K, ๐พ๐‘ = 1.85 for the reaction SO2(๐‘”) + 12 O2(๐‘”) โ‡Œ SO3(๐‘”) (c) What is the value of ๐พ๐‘ for the reaction in part (b)?

1089
views
๊ต๊ณผ์„œ ์งˆ๋ฌธ

The following equilibria were attained at 823 K:

CoO(s) + H2(g) โ†’ Co(s) + H2O(g) Kc = 67

CoO(s) + CO(g) โ†’ Co(s) + CO2(g) Kc = 490

Based on these equilibria, calculate the value of ๐พ๐‘ for H2(๐‘”)+ CO2(๐‘”) โ‡Œ CO(๐‘”) + H2O(๐‘”) at 823 K.

๊ต๊ณผ์„œ ์งˆ๋ฌธ

Consider the following equilibrium, for which at ๐พ๐‘ = 0.0752 at 480ยฐC: 2 Cl2(๐‘”) + 2 H2O(๐‘”) โ‡Œ 4 HCl(๐‘”) + O2(๐‘”) (a) What is the value of ๐พ๐‘ for the reaction 4 HCl(๐‘”) + O2(๐‘”) โ‡Œ 2 Cl2(๐‘”) + 2 H2O(๐‘”)?

1
views