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 $\left[\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{CO}(g)\right] ?$

Short Answer

Expert verified
The activation energy for the reverse reaction is \(-91 \, \mathrm{kJ/mol}\).

Step by step solution

01

Identify the given values

For this exercise, we are given the activation energy for the forward reaction (125 kJ/mol), the change in energy of the reaction (-216 kJ/mol), and we are asked to find the activation energy for the reverse reaction.
02

Use the formula

To find the activation energy for the reverse reaction, we should use the formula that relates the activation energy of the forward and reverse reactions and the change in energy of the reaction: Activation energy of reverse reaction (Ea_reverse) = Activation energy of forward reaction (Ea_forward) + Change in energy of reaction (∆E) Plug in the given values into the formula: Ea_reverse = Ea_forward + ∆E Ea_reverse = \(125 \, \mathrm{kJ/mol} - 216 \, \mathrm{kJ/mol}\)
03

Calculate the activation energy for the reverse reaction

To find the activation energy for the reverse reaction, simply perform the subtraction in the equation: Ea_reverse = \(-91 \, \mathrm{kJ/mol}\) The activation energy for the reverse reaction is -91 kJ/mol. This negative value indicates that the activation energy barrier for the reverse reaction is actually lower than that of the forward reaction, making the reverse reaction more favorable to occur under certain conditions.

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Most popular questions from this chapter

Consider the reaction $$ 3 \mathrm{A}+\mathrm{B}+\mathrm{C} \longrightarrow \mathrm{D}+\mathrm{E} $$ where the rate law is defined as $$ -\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}] $$ An experiment is carried out where $[\mathrm{B}]_{0}=[\mathrm{C}]_{0}=1.00 \mathrm{M}$ and \([\mathrm{A}]_{0}=1.00 \times 10^{-4} \mathrm{M}\) a. If after \(3.00 \min ,[\mathrm{A}]=3.26 \times 10^{-5} M,\) calculate the value of \(k .\) b. Calculate the half-life for this experiment. c. Calculate the concentration of \(B\) and the concentration of A after 10.0 min.

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 \(\mathrm{kJ} / \mathrm{mol} .\) In the presence of a catalyst at $37^{\circ} \mathrm{C}\( the rate constant for the reaction increases by a factor of \)2.50 \times 10^{3}$ as compared with the uncatalyzed reaction. Assuming the frequency factor \(A\) is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

What are the units for each of the following if the concentrations are expressed in moles per liter and the time in seconds? a. rate of a chemical reaction b. rate constant for a zero-order rate law c. rate constant for a first-order rate law d. rate constant for a second-order rate law e. rate constant for a third-order rate law

Consider the hypothetical reaction $$ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} $$ where the rate law is $$ \text {Rate} =-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}][\mathrm{B}]^{2} $$ An experiment is carried out where \([\mathrm{A}]_{0}=1.0 \times 10^{-2} M\) \([\mathrm{B}]_{0}=3.0 M,\) and \([\mathrm{C}]_{0}=2.0 M .\) The reaction is started, and after 8.0 seconds, the concentration of \(\mathrm{A}\) is $3.8 \times 10^{-3} \mathrm{M}$ a. Calculate the value of k for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after 13.0 seconds. d. Calculate the concentration of C after 13.0 seconds.

The rate law for the reaction $$ \begin{array}{c}{\mathrm{Cl}_{2}(g)+\mathrm{CHCl}_{3}(g) \longrightarrow \mathrm{HCl}(g)+\mathrm{CCl}_{4}(g)} \\ {\text { Rate }=k\left[\mathrm{Cl}_{2}\right]^{1 / 2}\left[\mathrm{CHCl}_{3}\right]}\end{array} $$ What are the units for \(k,\) assuming time in seconds and concentration in mol/L?

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