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Chapter 13: Problem 38

Given the same reactant concentrations, the reaction $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g) $$ at \(250^{\circ} \mathrm{C}\) is \(1.50 \times 10^{3}\) times as fast as the same reaction at \(150^{\circ} \mathrm{C}\). Calculate the activation energy for this reaction. Assume that the frequency factor is constant.

Short Answer

Expert verified
After plugging the given values and R (8.314 J/mol K), we find the activation energy, Ea = 132 kJ/mol

Step by step solution

01

Convert temperatures to Kelvin

In order to use the Arrhenius equation we need to convert the given temperatures to Kelvin: \( T_1 = 150^\circ C + 273.15 = 423.15 K \) and \( T_2 = 250^\circ C + 273.15 = 523.15 K \)
02

Express the ratio of rate constants

The problem mentions that the reaction at \(250^{\circ} \mathrm{C}\) (represented by \(k_2\)) is \(1.50 \times 10^{3}\) times as fast as the same reaction at \(150^{\circ} \mathrm{C}\) (represented by \(k_1\)). This implies \( k_2/k_1 = 1.50 \times 10^{3} \).
03

Set up the Arrhenius equations for k1 and k2

Write out the Arrhenius equations for \(k_1\) and \(k_2\): \(k_1 = A \exp(-E_a / (R \cdot T_1))\) and \(k_2 = A \exp(-E_a / (R \cdot T_2))\)
04

Solve for the Activation Energy (Ea)

Substitute \(k_2 / k_1\) from step 2 into the equation and solve for the Activation Energy (Ea): Here we divide the Arrhenius equation for \(k_2\) by the equation for \(k_1\), to get \(k_2 / k_1 = \exp(E_a/R (1/T_1 - 1/T_2))\). We can then take natural logarithm on both sides to get \(\ln(k_2 / k_1) = E_a/R (1/T_1 - 1/T_2)\). Solving for \(E_a\) gives \(E_a = R \ln(k_2 / k_1) / (1/T_1 - 1/T_2)\)

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Most popular questions from this chapter

The reaction of \(\mathrm{G}_{2}\) with \(\mathrm{E}_{2}\) to form \(2 \mathrm{EG}\) is exothermic, and the reaction of \(\mathrm{G}_{2}\) with \(\mathrm{X}_{2}\) to form \(2 \mathrm{XG}\) is endothermic. The activation energy of the exothermic reaction is greater than that of the endothermic reaction. Sketch the potential energy profile diagrams for these two reactions on the same graph.

When \(6 \mathrm{~g}\) of granulated \(\mathrm{Zn}\) is added to a solution of \(2 M \mathrm{HCl}\) in a beaker at room temperature, hydrogen gas is generated. For each of the following changes (at constant volume of the acid) state whether the rate of hydrogen gas evolution will be increased, decreased, or unchanged: (a) \(6 \mathrm{~g}\) of powdered \(\mathrm{Zn}\) is used; (b) \(4 \mathrm{~g}\) of granulated Zn is used; (c) \(2 \mathrm{M}\) acetic acid is used instead of \(2 M \mathrm{HCl} ;\) (d) temperature is raised to \(40^{\circ} \mathrm{C}\)

To prevent brain damage, a drastic medical procedure is to lower the body temperature of someone who has suffered cardiac arrest. What is the physiochemical basis for this treatment?

The rate law for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g) $$ is given by rate \(=k[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right] .\) (a) What is the order of the reaction? (b) A mechanism involving the following steps has been proposed for the reaction: $$ \begin{array}{c} \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}_{2}(g) \\ \mathrm{NOCl}_{2}(g)+\mathrm{NO}(g) \longrightarrow 2 \mathrm{NOCl}(g) \end{array} $$ If this mechanism is correct, what does it imply about the relative rates of these two steps?

Determine the overall orders of the reactions to which the following rate laws apply: (a) rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\), (b) rate \(=k,(\mathrm{c})\) rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]^{\frac{1}{2}}\) (d) rate \(=\) \(k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\)
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