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Chapter 11: Problem 69

The activation energy for the decomposition of \(\mathrm{HI}(g)\) to \(\mathrm{H}_{2}(g)\) and \(\mathrm{I}_{2}(g)\) is \(186 \mathrm{kJ} / \mathrm{mol} .\) The rate constant at \(555 \space\mathrm{K}\) is \(3.52 \times\) \(10^{-7} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s} .\) What is the rate constant at \(645 \space\mathrm{K} ?\)

Short Answer

Expert verified
The rate constant at 645 K for the decomposition of HI(g) to H2(g) and I2(g) is approximately \(1.99 \times 10^{-5} L/mol·s\).

Step by step solution

01

Write down the Arrhenius equation and the given values

The Arrhenius equation relates the activation energy, temperature, and rate constant: \[ k = Ae^{-\frac{E_a}{RT}} \] Where: - \(k\) is the rate constant - \(A\) is the pre-exponential factor or frequency factor - \(E_a\) is the activation energy - \(R\) is the gas constant (8.314 J/mol·K) - \(T\) is the temperature in Kelvin We are given: - Activation energy, \(E_a = 186 \space kJ/mol = 186,000 J/mol\) - Rate constant at 555 K, \(k_1 = 3.52 \times 10^{-7} L/mol·s\) - Temperature at 555 K, \(T_1 = 555 K\) - Temperature at 645 K, \(T_2 = 645 K\) We want to find the rate constant at 645 K, \(k_2\).
02

Divide the Arrhenius equation for the two different temperatures

By taking the ratio of the Arrhenius equation at 645 K and 555 K, we can eliminate the pre-exponential factor \(A\): \[\frac{k_2}{k_1} = \frac{Ae^{-\frac{E_a}{R T_2}}}{Ae^{-\frac{E_a}{R T_1}}}\]
03

Simplify the equation and solve for \(k_2\)

Simplify the equation in Step 2: \[\frac{k_2}{k_1} = \frac{e^{-\frac{E_a}{R T_2}}}{e^{-\frac{E_a}{R T_1}}}\] Now take the natural logarithm of both sides: \[ln(\frac{k_2}{k_1}) = ln(\frac{e^{-\frac{E_a}{R T_2}}}{e^{-\frac{E_a}{R T_1}}}) = ln(e^{-\frac{E_a}{R T_2}})-ln(e^{-\frac{E_a}{R T_1}})\] Using the property of logarithms to rewrite the equation again: \[-\frac{E_a}{R T_2} + \frac{E_a}{R T_1} = ln(\frac{k_2}{k_1})\] Solve for \(k_2\): \[k_2 = k_1 \cdot e^{\frac{-E_a}{R T_2} + \frac{E_a}{R T_1}}\]
04

Insert the given values and calculate \(k_2\)

Insert the given values into the equation from Step 3 and calculate \(k_2\): \[k_2 = (3.52 \times 10^{-7} L/mol·s) \cdot e^{\frac{-186,000 J/mol}{(8.314 J/mol·K)(645 K)} + \frac{186,000 J/mol}{(8.314 J/mol·K)(555 K)}}\] \[k_2 \approx 1.99 \times 10^{-5} L/mol·s\] So the rate constant at 645 K is approximately \(1.99 \times 10^{-5} L/mol·s\).

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Most popular questions from this chapter

The activation energy for the reaction $$\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)$$ is \(125 \mathrm{kJ} / \mathrm{mol},\) and \(\Delta E\) for the reaction is \(-216 \mathrm{kJ} / \mathrm{mol}\). What is the activation energy for the reverse reaction \(\left[\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{CO}(g)\right] ?\)

The reaction $$A \longrightarrow B+C$$ is known to be zero order in A and to have a rate constant of \(5.0 \times 10^{-2} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C} .\) An experiment was run at \(25^{\circ} \mathrm{C}\) where \([\mathrm{A}]_{0}=1.0 \times 10^{-3} \mathrm{M}\) a. Write the integrated rate law for this reaction. b. Calculate the half-life for the reaction. c. Calculate the concentration of \(\mathrm{B}\) after \(5.0 \times 10^{-3} \mathrm{s}\) has elapsed assuming \([\mathrm{B}]_{0}=0\)

One reason suggested for the instability of long chains of silicon atoms is that the decomposition involves the transition state shown below: The activation energy for such a process is \(210 \space\mathrm{kJ} / \mathrm{mol}\), which is less than either the \(\mathrm{Si}-\mathrm{Si}\) or the \(\mathrm{Si}-\mathrm{H}\) bond energy. Why would a similar mechanism not be expected to play a very important role in the decomposition of long chains of carbon atoms as seen in organic compounds?

Would the slope of a \(\ln (k)\) versus \(1 / T\) plot (with temperature in kelvin) for a catalyzed reaction be more or less negative than the slope of the \(\ln (k)\) versus \(1 / T\) plot for the uncatalyzed reaction? Explain. Assume both rate laws are first-order overall.

Assuming that the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section \(11-7\) is correct, would you predict that the product of the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) would be \(\mathrm{CH}_{2} \mathrm{D}-\mathrm{CH}_{2} \mathrm{D}\) or \(\mathrm{CHD}_{2}-\mathrm{CH}_{3} ?\) How could the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) be used to confirm the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section \(11-7 ?\)
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