Reaction rates and Pressure? Check my reasoning!

asked by @jankj1 • almost 2 years ago • Chemistry • 5 pts


When will the reaction be slower? A. Lower temperatures B. Smaller Volumes C. Higher Pressures D. With a Catalyst E. None of the above

I know it's not D or E, but the other 3 are techically fair game, since it's a gaseous Reaction, correct? The answer given is "A," but a smaller volume or higher pressure would force the equilibrium to the side with less moles, which is the reactants. Is my reasoning correct

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3 answers

Hey Jank! In this question we are not provided with any kind of information that describes that equilibrium of the reaction is prone to shift towards either left or right side of the equation. Therefore, option B & C can also be excluded. Secondly, even if that information was provided we could not go for option B & C because lowering volume and increasing pressure will move the reaction towards its favourable side. That means if equilibrium was shifted more towards products then these conditions will definitely let the reactants to collide more frequently and eventually increase the speed of reaction rather slowing it down.

We can rationalise option A in such a way that lowering temperature will decrease the kinetic energy of the molecules so they will be colliding less frequently and the speed of the reaction will be slower.

Hope this makes sense :-)

answered by @abdullah • almost 2 years ago

@abdullah, so If I was to change the pressure and volume, would it make a difference whatsoever in the rate of the reaction?

answered by @jankj1 • almost 2 years ago

Hey! So for this problem, we're dealing with Chemical Kinetics, not equilibrium. According to the equation shown, this reaction is going in a single direction, meaning equilibrium is not present. Since equilibrium is not present, choices B and C can be disregarded since they deal with equilibrium. Actually, you are able to tell the direction of the reaction if this was an equilibrium problem. Remember that decreasing volume and increasing pressure leads to a shift towards the side with the least amount of moles. Again, this is only if the reaction was at equilibrium, but it's not so we disregard B and C.

We're left with A, D, and E. Understand that catalysts are used to increase the rate of the reaction. Here, we're looking to decrease the rate of the reaction, so we can disregard D as well. Finally, A makes sense because we know that increasing the temperature of a reaction will make the reaction faster. Higher temperatures make the molecules move faster, thus creating a faster reaction. So, decreasing the temperature will also decrease the rate of the reaction. The correct answer is A. Hope this helped!

answered by @sabrina • almost 2 years ago