The gas-phase reaction 2 N2 O5(g) ⟶4 NO2(g)+O2(g) has an activation energy of 103 kJ, and the ra

The gas-phase reaction 2 N2 O5(g) ⟶4 NO2(g)+O2(g) has an...

The gas-phase reaction $$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ has an activation energy of $103 \mathrm{kJ},$ and the rate constant is $0.0900 \min ^{-1}$ at $328.0 \mathrm{K}$. Find the rate constant at $318.0 \mathrm{K}$

The gas-phase reaction
$$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$
has an activation energy of $103 \mathrm{kJ},$ and the rate constant is $0.0900 \min ^{-1}$ at $328.0 \mathrm{K}$. Find the rate constant at $318.0 \mathrm{K}$

Key Concepts

Activation Energy

Activation energy is the minimum amount of energy required for a chemical reaction to occur. It plays a critical role in determining how fast a reaction proceeds; a higher energy barrier slows down the reaction rate because fewer molecules have sufficient energy to overcome it at a given temperature.

Rate Constant

The rate constant is a proportionality constant in the rate law that links the rate of a reaction to the concentrations of the reactants. It is specific to a reaction at a given temperature and embodies the effects of factors such as activation energy and frequency of effective collisions.

Arrhenius Equation

The Arrhenius equation describes the temperature dependence of the rate constant by expressing how the rate changes with temperature and activation energy. It shows that even small changes in temperature can significantly alter the rate constant, reflecting the exponential relationship between temperature and reaction rate.

Temperature Dependence of Reaction Rates

The temperature dependence of reaction rates is a fundamental concept in chemical kinetics that explains how reactions typically proceed faster at higher temperatures. This is due to the increased kinetic energy of molecules, which enhances the probability that collisions will overcome the activation energy barrier.

Transcript

00:01 So there's a couple of ways you can solve a problem where you're given rate constants, activation energies, and temperatures, and asked to solve for them.

00:08 They're both fundamentally based off the erratus equation, which is that k equals ae, it's the negative ea over rt, where a is a constant for a reaction.

00:21 And what people have done is they've basically derived out from this equation, something that would work for multiple equations, which is that the natural.

00:30 Log of k2 over k1, where these subscripts refer to two different temperatures, equals the negative activation energy over the gas constant times one over one temperature, one over the other one.

00:46 And so you could plug in everything we don't, everything we have, and the thing that we would be missing would be one of these rate constants and we could solve for it.

00:56 I'm going to present another way, which is you first solve this with the information we have, to find a and then re -plug it and so i'll rewrite what we have is k equals a eats the negative ea over rt and so what we have is k equals 0 .09 inverse minutes a is what we're going to look for and it's going to be e to the negative 103 kilojoules but r is going to be in joules per mole kelvin so turn this into joules r is 8 .314 the temperature is 324 the temperature is 320 quentin note about what r is to be more specific 8 .314 joules per mole kelvin this is the gas constant we use and you do this you find an absolutely nasty number for a it is equal to two two seven nine two one eight two five one eight six two six nine four inverse minutes and so that sucks but you just keep in your calculator and and now we plug in again to find the rate constant at a different times.

02:13 So remember, k equals a, definitely not rewriting that.

02:17 Just plug that in...

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