Increasing the surface area will increase the reaction rate in:
1. N2(g) + O2(g) -> 2NO(g)
2. 2Mg(s) + O2(g) -> 2MgO(s)
3. CaCO3(s) + 2H+(aq) -> Ca2+(aq) + H2O(l) + CO2(g)
MY WORK: I think it is 2 & possibly 3. 1 is a homogeneous reaction but is 3 a heterogeneous reaction? gas and aqueous?
I don't see a reaction 3a. I see a reaction 3 and that one heterogeneous. So is 2. I think the reaction rate will increase for BOTH 2 and 3 although I think 3 is less so than 2.
oh I mean, is 3 a heterogeneous reaction, not 3a. But a gas and an aqueous is a heterogeneous reaction correct?
Thanks!
Yes, 3 is a heterogeneous reaction. So is 2.
thank you very much!
You're correct that reaction 1 is a homogeneous reaction, where all reactants and products are in the same phase (gaseous state). Reaction 3 is indeed a heterogeneous reaction, as it involves the reaction between a solid (CaCO3) and substances in different phases (aqueous H+ and gaseous CO2).
Regarding the effect of surface area on reaction rate, let's analyze each reaction:
1. N2(g) + O2(g) -> 2NO(g)
In this reaction, increasing the surface area wouldn't affect the reaction rate since all the reactants and products are in the gaseous phase, and the surface area of gases is a constant. Therefore, the surface area does not play a significant role in this reaction.
2. 2Mg(s) + O2(g) -> 2MgO(s)
In this reaction, since one of the reactants is a solid (Mg), increasing its surface area by using smaller pieces or powder will indeed increase the reaction rate. This is because a larger surface area provides more opportunities for the gas molecules (in this case, O2) to come into contact with the solid surface (Mg), leading to more frequent and effective collisions and faster reaction rate.
3. CaCO3(s) + 2H+(aq) -> Ca2+(aq) + H2O(l) + CO2(g)
Similar to reaction 2, increasing the surface area of the solid reactant (CaCO3) in this reaction will also increase the reaction rate. More surface area means that more solid particles are exposed to aqueous H+ ions, allowing for more collisions and a faster reaction rate.
So, the correct answer is indeed options 2 and 3. Increasing the surface area will increase the reaction rate in reactions 2 and 3, where one or more reactants are in the solid phase.