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Chapter 15: Problem 99

The "boat" form and "chair" form of cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) interconverts as shown here: In this representation, the \(\mathrm{H}\) atoms are omitted and a \(\mathrm{C}\) atom is assumed to be at each intersection of two lines (bonds). The conversion is first order in each direction. The activation energy for the chair \(\longrightarrow\) boat conversion is \(41 \mathrm{~kJ} / \mathrm{mol}\). If the frequency factor is \(1.0 \times 10^{12} \mathrm{~s}^{-1}\), what is \(k_{1}\) at \(298 \mathrm{~K}\) ? The equilibrium constant \(K_{c}\) for the reaction is \(9.83 \times 10^{3}\) at \(298 \mathrm{~K}\).

Short Answer

Expert verified
Use a calculator to solve for \(k\), which will be in \(s^{-1}\).

Step by step solution

01

Write down the Arrhenius equation

The Arrhenius equation is given by \( k = A \exp\left(-\frac{E_a}{RT}\right) \) where \(k\) is the rate constant, \(A\) is the frequency factor, \(E_a\) is the activation energy, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
02

Substitute the given values into the equation

Substitute the given values \(A = 1.0 \times 10^{12} s^{-1}\), \(E_a = 41 kJ/mol = 41000 J/mol\), \(R = 8.314 J/(mol \cdot K)\), and \(T = 298 K\) into the equation to get \(k = 1.0 \times 10^{12} s^{-1} \exp\left(-\frac{41000 J/mol}{8.314 J/(mol \cdot K) \cdot 298 K}\right)\)
03

Calculate the value of the rate constant

Calculate the value of the rate constant \(k\). This is achieved through performing the arithmetic operation to get a numerical value for the rate constant \(k_1\).

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Most popular questions from this chapter

Briefly describe the importance of equilibrium in the study of chemical reactions.

When dissolved in water, glucose (corn sugar) and fructose (fruit sugar) exist in equilibrium as follows: fructose \(\rightleftharpoons\) glucose A chemist prepared a \(0.244 M\) fructose solution at \(25^{\circ} \mathrm{C}\). At equilibrium, it was found that its concentration had decreased to \(0.113 M\). (a) Calculate the equilibrium constant for the reaction. (b) At equilibrium, what percentage of fructose was converted to glucose?

A quantity of 0.20 mole of carbon dioxide was heated at a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Under this condition, the average molar mass of the gases was found to be \(35 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\). (b) What is the \(K_{P}\) for the equilibrium if the total pressure was 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.)

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is \(5.60 \times 10^{4}\) at \(350^{\circ} \mathrm{C} . \mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are mixed initially at 0.350 atm and 0.762 atm, respectively, at \(350^{\circ} \mathrm{C}\). When the mixture equilibrates, is the total pressure less than or greater than the sum of the initial pressures, 1.112 atm?

Consider the heterogeneous equilibrium process: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ At \(700^{\circ} \mathrm{C}\), the total pressure of the system is found to be 4.50 atm. If the equilibrium constant \(K_{P}\) is 1.52 , calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\)
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