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For the reaction: $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ the frequency factor \(A\) is \(8.7 \times 10^{12} \mathrm{~s}^{-1}\) and the activation energy is \(63 \mathrm{~kJ} / \mathrm{mol}\). What is the rate constant for the reaction at \(75^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The rate constant at 75°C is approximately 3.12 s⁻¹.

Step by step solution

01

Convert Temperature to Kelvin

First, convert the temperature from degrees Celsius to Kelvin. The formula is: \[ T(K) = T(°C) + 273.15 \] Substitute the given temperature into the formula:\[ T(K) = 75 + 273.15 = 348.15 \] So, the temperature is 348.15 K.
02

Write the Arrhenius Equation

The Arrhenius equation gives the relationship between the rate constant \( k \) and the temperature \( T \), as follows:\[ k = A e^{-E_a / (RT)} \] where:- \( A \) is the frequency factor,- \( E_a \) is the activation energy,- \( R \) is the gas constant (8.314 J/mol K),- \( T \) is the temperature in Kelvin.
03

Convert Activation Energy

Convert the activation energy from kJ/mol to J/mol because the gas constant \( R \) is in J/mol K. The conversion is:\[ E_a = 63 \text{ kJ/mol} \times 1000 = 63000 \text{ J/mol} \] Now, \( E_a = 63000 \text{ J/mol} \).
04

Substitute Values into Arrhenius Equation

Substitute the given values and constants into the Arrhenius equation:\[ k = 8.7 \times 10^{12} \text{ s}^{-1} \times \exp\left(-\frac{63000}{8.314 \times 348.15}\right) \]
05

Evaluate the Exponent

Calculate the exponent:\[ \frac{63000}{8.314 \times 348.15} \approx 21.740 \]
06

Calculate the Rate Constant

Now calculate the exponential and the rate constant:\[ k = 8.7 \times 10^{12} \text{ s}^{-1} \times \exp(-21.740) \] Using a calculator for \( \exp(-21.740) \), we find:\[ \exp(-21.740) \approx 3.59 \times 10^{-10} \] So,\[ k \approx 8.7 \times 10^{12} \times 3.59 \times 10^{-10} \] \[ k \approx 3.12 \text{ s}^{-1} \]
07

Final Step: Review and Conclude

Review the calculations for any sign of miscalculation or misplacement of decimal points. The final result appears consistent with typical calculations of rate constants using the Arrhenius equation.

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Key Concepts

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

Activation Energy
Activation energy is a crucial concept in reaction kinetics, helping to explain why some reactions are faster or slower than others. It represents the minimum energy required to initiate a chemical reaction. This energy is needed to break the chemical bonds in the reactants so that new bonds can form in the products. In the Arrhenius equation, the activation energy is denoted by \(E_a\). The unit of activation energy is often expressed in kilojoules per mole (kJ/mol). In many calculations, it is necessary to convert this value to joules per mole (J/mol) because constants like the gas constant \(R\) are given in those units. To do this conversion, simply multiply by 1,000 (since 1 kJ = 1,000 J). Understanding activation energy helps in predicting the rate at which a reaction will occur, which is essential for controlling and optimizing chemical processes in both industrial and laboratory settings.
Rate Constant Calculation
Calculating the rate constant of a chemical reaction gives insight into the speed of the reaction under specific conditions. This rate constant, denoted by \(k\), can be calculated using the Arrhenius equation:\[ k = A e^{-E_a / (RT)} \] Here:
  • \(A\) represents the frequency factor, which accounts for the frequency of collisions between reactant molecules.
  • \(E_a\) is the activation energy.
  • \(R\) is the universal gas constant, which is approximately 8.314 J/mol K.
  • \(T\) stands for the temperature in Kelvin.
The exponential part \(e^{-E_a / (RT)}\) shows how the rate constant decreases exponentially with increasing activation energy and increases with temperature. A higher rate constant means a faster reaction at a given temperature. To solve for \(k\), substitute the known values into the equation, carefully calculate the exponent, and multiply by the frequency factor. Whether you're working on homework, a lab experiment, or an industrial-scale production, understanding this concept is vital for precise reaction rate predictions.
Temperature Conversion
In chemical kinetics, temperature conversion is fundamental because temperature significantly affects reaction rates. The Arrhenius equation requires the temperature to be in Kelvin to ensure consistency in units. To convert from Celsius to Kelvin, use the formula:\[ T(K) = T(°C) + 273.15 \]This conversion is straightforward but essential. Temperature in Kelvin accounts for the absolute thermal kinetic energy present, providing a true measure of the energy available to overcome the activation energy barrier. For example, a temperature of 75°C converts to:
  • 75 + 273.15 = 348.15 K
This accurate conversion is needed to ensure that calculations involving the Arrhenius equation and other kinetic formulas yield reliable results. Proper understanding and conversion of temperature units help prevent errors in experiments and chemical analysis, allowing for more accurate predictions of reaction behavior.

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Most popular questions from this chapter

A factory that specializes in the refinement of transition metals such as titanium was on fire. The firefighters were advised not to douse the fire with water. Why?

Radioactive plutonium- \(239\left(t_{1 / 2}=2.44 \times 10^{5} \mathrm{yr}\right)\) is used in nuclear reactors and atomic bombs. If there are \(5.0 \times 10^{2} \mathrm{~g}\) of the isotope in a small atomic bomb, how long will it take for the substance to decay to \(1.0 \times 10^{2} \mathrm{~g}\), too small an amount for an effective bomb?

To carry out metabolism, oxygen is taken up by hemoglobin \((\mathrm{Hb})\) to form oxyhemoglobin \(\left(\mathrm{Hb} \mathrm{O}_{2}\right)\) according to the simplified equation: $$ \mathrm{Hb}(a q)+\mathrm{O}_{2}(a q) \stackrel{k}{\longrightarrow} \mathrm{HbO}_{2}(a q) $$ where the second-order rate constant is \(2.1 \times 10^{6} / M \cdot \mathrm{s}\) at \(37^{\circ} \mathrm{C}\). For an average adult, the concentrations of \(\mathrm{Hb}\) and \(\mathrm{O}_{2}\) in the blood at the lungs are \(8.0 \times 10^{-6} \mathrm{M}\) and \(1.5 \times 10^{-6} M,\) respectively. (a) Calculate the rate of formation of \(\mathrm{HbO}_{2}\). (b) Calculate the rate of consumption of \(\mathrm{O}_{2}\). (c) The rate of formation of \(\mathrm{HbO}_{2}\) increases to \(1.4 \times 10^{-4} M / \mathrm{s}\) during exercise to meet the demand of the increased metabolism rate. Assuming the \(\mathrm{Hb}\) concentration to remain the same, what must the oxygen concentration be to sustain this rate of \(\mathrm{HbO}_{2}\) formation?

Explain what is meant by the rate law of a reaction.

Define half-life. Write the equation relating the half-life of a first-order reaction to the rate constant.

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