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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 25a,b
15์žฅ, ๋ฌธ์ œ 25a,b

At 1000 K, ๐พ๐‘ = 1.85 for the reaction SO2(๐‘”) + 1/2 O2(๐‘”) โ‡Œ SO3(๐‘”) (a) What is the value of Kp for the reaction SO3(๐‘”) โ‡Œ SO2(๐‘”) + 1/2 O2(๐‘”)? (b) What is the value of Kp for the reaction 2 SO2(๐‘”) + O2(๐‘”) โ‡Œ 2 SO3(๐‘”)?

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1
To find the value of Kp for the reverse reaction SO3(g) โ‡Œ SO2(g) + 1/2 O2(g), we need to take the reciprocal of the given Kp for the forward reaction. This is because the equilibrium constant for a reverse reaction is the inverse of the equilibrium constant for the forward reaction.
For part (a), calculate the reciprocal of Kp: Kp(reverse) = 1 / Kp(forward). Substitute the given Kp value of 1.85 into this expression.
For part (b), we need to determine the Kp for the reaction 2 SO2(g) + O2(g) โ‡Œ 2 SO3(g). Notice that this reaction is the original reaction multiplied by 2.
When a chemical equation is multiplied by a factor, the equilibrium constant is raised to the power of that factor. Therefore, for the reaction 2 SO2(g) + O2(g) โ‡Œ 2 SO3(g), Kp(new) = (Kp(original))^2.
Substitute the given Kp value of 1.85 into the expression for Kp(new) to find the equilibrium constant for the reaction 2 SO2(g) + O2(g) โ‡Œ 2 SO3(g).

์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 (Kp)

The equilibrium constant, Kp, is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. It is calculated using the partial pressures of gases involved in the reaction. A larger Kp value indicates a greater concentration of products at equilibrium, while a smaller value suggests a higher concentration of reactants.
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Equilibrium Constant Expressions

Reaction Quotient (Q)

The reaction quotient, Q, is a measure of the relative amounts of products and reactants present in a reaction at any point in time, not necessarily at equilibrium. It is calculated in the same way as Kp but uses the current concentrations or partial pressures. Comparing Q to Kp helps predict the direction in which a reaction will proceed to reach equilibrium.
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Reaction Quotient Q

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 for related reactions.
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Le Chatelier's Principle
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๊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)?

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๊ต๊ณผ์„œ ์งˆ๋ฌธ

Consider the following equilibrium: 2 H2(g) + S2(g) โ‡Œ 2 H2S(g) Kc = 1.08 ร— 107 at 700ยฐC (c) Calculate the value of ๐พ๐‘ if you rewrote the equation H2(g) + 1/2 S2(g) โ‡Œ H2S(g)

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๊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: 2 H2(g) + S2(g) โ‡Œ 2 H2S(g) Kc = 1.08 ร— 107 at 700ยฐC (b) Does the equilibrium mixture contain mostly H2 and S2 or mostly H2S?

๊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‘”)?

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