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Chapter 14: Problem 51

The rate constant for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow\) \(2 \mathrm{HI}(\mathrm{g})\) has been determined at the following temperatures: \(599 \mathrm{K}, k=5.4 \times 10^{-4} \mathrm{M}^{-1} \mathrm{s}^{-1} ; 683 \mathrm{K}, k=2.8 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Calculate the activation energy for the reaction.

Short Answer

Expert verified
The activation energy for the given reaction is approximately \(94.23 \, kJ/mol\).

Step by step solution

01

Write down the Arrhenius equation

We first write down the Arrhenius equation: \( k = Ae^{-Ea/RT}\).
02

Insert the given values and set up two equations

We now insert the given values into the Arrhenius equation. We get two equations: \(5.4*10^{-4} = Ae^{-Ea/(8.314*599)}\) and \(2.8*10^{-2} = Ae^{-Ea/(8.314*683)}\). Here we have used R = 8.314 J/(K*mol).
03

Divide the two equations

We divide the first equation by the second to get rid of A: \( \frac{5.4*10^{-4}}{2.8*10^{-2}} = \frac{e^{-Ea/(8.314*599)}}{e^{-Ea/(8.314*683)}} \)
04

Simplify and solve for Ea

This simplifies to \(0.0193 = e^{Ea/(8.314)*(1/683-1/599)}\). We then take the natural logarithm of both sides and solve for Ea: \(Ea= -8.314*ln(0.0193)/(1/683-1/599)\).
05

Calculate Ea

Plugging in the numbers gives us \(Ea=93830.77 \, J/mol \) or \(94.23\,kJ/mol\).

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Most popular questions from this chapter

A reaction is \(50 \%\) complete in 30.0 min. How long after its start will the reaction be \(75 \%\) complete if it is (a) first order; (b) zero order?

Certain gas-phase reactions on a heterogeneous catalyst are first order at low gas pressures and zero order at high pressures. Can you suggest a reason for this?

For the first-order reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ \(t_{1 / 2}=22.5 \mathrm{h}\) at \(20^{\circ} \mathrm{C}\) and \(1.5 \mathrm{h}\) at \(40^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) If the Arrhenius constant \(A=2.05 \times 10^{13} \mathrm{s}^{-1}\) determine the value of \(k\) at \(30^{\circ} \mathrm{C}\).

Derive a plausible mechanism for the following reaction in aqueous solution, \(\mathrm{Hg}_{2}^{2+}+\mathrm{Tl}^{3+} \longrightarrow 2 \mathrm{Hg}^{2+}+\mathrm{Tl}^{+}\) for which the observed rate law is: rate \(=k\left[\mathrm{Hg}_{2}^{2+1}\right]\) \(\left.[\mathrm{T}]^{3+}\right] /\left[\mathrm{Hg}^{2+}\right].\)

In the first-order reaction \(A \longrightarrow\) products, it is found that \(99 \%\) of the original amount of reactant \(A\) decomposes in 137 min. What is the half-life, \(t_{1 / 2}\), of this decomposition reaction?
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