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Ch.14 - Chemical Kinetics
Brown - Chemistry: The Central Science 15th Edition
Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970당신이 사용하는 게 아니라요?교과서 변경
14장, 문제 118d

The reaction between ethyl iodide and hydroxide ion in ethanol (C2H5OH) solution, C2H5I(alc) + OH-(alc) → C2H5OH(l) + I-(alc), has an activation energy of 86.8 kJ/mol and a frequency factor of 2.10 × 1011 M-1 s-1. (d) Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at 50 C.

검증된 단계별 안내
1
Convert the temperature from Celsius to Kelvin by adding 273.15 to the given temperature (50 °C).
Use the Arrhenius equation: \( k = A \cdot e^{-\frac{E_a}{RT}} \), where \( k \) is the rate constant, \( A \) is the frequency factor, \( E_a \) is the activation energy, \( R \) is the gas constant (8.314 J/mol·K), and \( T \) is the temperature in Kelvin.
Substitute the given values into the Arrhenius equation: \( A = 2.10 \times 10^{11} \text{ M}^{-1} \text{ s}^{-1} \), \( E_a = 86.8 \text{ kJ/mol} \) (convert this to J/mol by multiplying by 1000), and \( T \) is the temperature in Kelvin calculated in step 1.
Calculate the exponent \( -\frac{E_a}{RT} \) using the values from step 3.
Compute the rate constant \( k \) by evaluating the Arrhenius equation with the calculated exponent and the given frequency factor.

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주요 개념

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

Arrhenius Equation

The Arrhenius equation relates the rate constant of a chemical reaction to temperature and activation energy. It is expressed as k = A * e^(-Ea/RT), where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the universal gas constant, and T is the temperature in Kelvin. This equation shows how an increase in temperature can lead to an increase in the rate constant, thereby accelerating the reaction.
추천 영상:
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01:20
Arrhenius Equation

Activation Energy

Activation energy (Ea) is the minimum energy required for a chemical reaction to occur. It represents the energy barrier that reactants must overcome to form products. In the context of the Arrhenius equation, a higher activation energy results in a lower rate constant at a given temperature, indicating that the reaction proceeds more slowly. Understanding activation energy is crucial for predicting reaction rates and mechanisms.
추천 영상:
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02:02
Activity Series Chart

Temperature Conversion

Temperature must be expressed in Kelvin when using the Arrhenius equation. To convert Celsius to Kelvin, add 273.15 to the Celsius temperature. For example, 50°C is equivalent to 323.15 K. Accurate temperature conversion is essential for correctly calculating the rate constant, as the rate constant is temperature-dependent and influences the overall reaction kinetics.
추천 영상:
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02:10
Temperature Conversion Example
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교과서 질문

The reaction between ethyl iodide and hydroxide ion in ethanol (C2H5OH) solution, C2H5I(alc) + OH-(alc) → C2H5OH(l) + I-(alc), has an activation energy of 86.8 kJ/mol and a frequency factor of 2.10 × 1011 M-1 s-1. (c) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion?

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교과서 질문

The mechanism for the oxidation of HBr by O2 to form 2 H2O and Br2 is shown in Exercise 14.74. (a) Calculate the overall standard enthalpy change for the reaction process.

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교과서 질문

The gas-phase reaction of NO with F2 to form NOF and F has an activation energy of Ea = 6.3 kJ>mol. and a frequency factor of A = 6.0 * 108 M-1 s-1. The reaction is believed to be bimolecular: NO1g2 + F21g2 ¡ NOF1g2 + F1g2 (e) Suggest a reason for the low activation energy for the reaction.

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교과서 질문

Enzymes are often described as following the two-step mechanism:

E + S  ⇌ ES (fast)

ES → E + P (slow)

where E = enzyme, S = substrate, ES = enzyme9substrate complex, and P = product.

(b) Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors. Write an additional elementary step to add into the preceding mechanism to account for the reaction of E with I, an inhibitor.

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교과서 질문

The mechanism for the oxidation of HBr by O2 to form 2 H2O and Br2 is shown in Exercise 14.74. (c) Draw a plausible Lewis structure for the intermediate HOOBr. To what familiar compound of hydrogen and oxygen does it appear similar?

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