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Ch.15 - Chemical Kinetics
Tro - Chemistry: A Molecular Approach 6th Edition
Tro6th EditionChemistry: A Molecular ApproachISBN: 9780137832217당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Tro 6th EditionCh.15 - Chemical Kinetics문제 122
15장, 문제 122

When HNO2 is dissolved in water, it partially dissociates according to the equation HNO2 ⇌ H+ + NO2-. A solution is prepared that contains 7.050 g of HNO2 in 1.000 kg of water. Its freezing point is -0.2929 °C. Calculate the fraction of HNO2 that has dissociated.

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Calculate the molality (m) of the solution using the formula: \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \). First, find the moles of HNO2 by dividing the mass of HNO2 (7.050 g) by its molar mass (47.013 g/mol). Then, divide the moles of HNO2 by the mass of water in kilograms (1.000 kg).
The freezing point depression (\( \Delta T_f \)) is given by the formula: \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant for water (1.86 °C kg/mol), and \( m \) is the molality. Since the freezing point is -0.2929 °C, \( \Delta T_f = 0.2929 \) °C.
Rearrange the freezing point depression formula to solve for \( i \): \( i = \frac{\Delta T_f}{K_f \cdot m} \). Substitute the known values of \( \Delta T_f \), \( K_f \), and \( m \) to find \( i \).
The van't Hoff factor \( i \) is related to the degree of dissociation \( \alpha \) by the equation: \( i = 1 + \alpha \cdot (n - 1) \), where \( n \) is the number of particles the solute dissociates into. For HNO2, \( n = 2 \) (HNO2 dissociates into H+ and NO2-). Solve for \( \alpha \) using the calculated \( i \).
The fraction of HNO2 that has dissociated is equal to the degree of dissociation \( \alpha \). Express \( \alpha \) as a percentage to find the fraction of HNO2 that has dissociated.>

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Dissociation of Acids

Dissociation refers to the process by which an acid separates into its constituent ions in solution. For weak acids like HNO2, this process is not complete, meaning only a fraction of the acid molecules dissociate into hydrogen ions (H+) and nitrite ions (NO2-). Understanding the degree of dissociation is crucial for calculating properties such as freezing point depression.
추천 영상:
가이드 코스
00:54
Acid-Base Dissociation Example

Freezing Point Depression

Freezing point depression is a colligative property that describes how the freezing point of a solvent decreases when a solute is added. The extent of this depression depends on the number of solute particles in solution, which can be calculated using the formula ΔTf = i * Kf * m, where 'i' is the van 't Hoff factor, 'Kf' is the freezing point depression constant, and 'm' is the molality of the solution.
추천 영상:
가이드 코스
01:59
Freezing Point Depression

Molality

Molality is a measure of concentration defined as the number of moles of solute per kilogram of solvent. It is expressed in moles per kilogram (mol/kg) and is particularly useful in colligative property calculations because it remains unaffected by temperature changes. In this problem, calculating the molality of the HNO2 solution is essential for determining the extent of freezing point depression and, subsequently, the fraction of dissociated HNO2.
추천 영상:
가이드 코스
00:40
Molality
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교과서 질문

Ethyl chloride vapor decomposes by the first-order reaction: C2H5Cl → C2H4 + HCl The activation energy is 249 kJ/mol, and the frequency factor is 1.6⨉1014 s-1. What fraction of the ethyl chloride decomposes in 15 minutes at this temperature?

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교과서 질문

Ethyl chloride vapor decomposes by the first-order reaction: C2H5Cl → C2H4 + HCl The activation energy is 249 kJ/mol, and the frequency factor is 1.6⨉1014 s-1. Find the value of the rate constant at 710 K.

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