- #1
RoboNerd
- 410
- 11
Hi everyone.
So I have been learning about equilibrium recently in class... and suppose I have the following reaction:
A(s) + B(aq) <-----> C(aq) + D(g)
[Note: in parentheses, I indicate the state of my substances]
So I was told recently that pulverizing my solid "A" would not change the position of equilibrium as nothing will happen... but I recalled our unit on Kinetics which says that increasing the surface area of my reactants will increase frequency of collisions which will increase the rate of the chemical reaction.
So, my impression is that pulverizing A would increase the surface area and increase the collisions for the forward reaction. Thus, the forward reaction would speed up, and a shift to the products would occur with [C] and [D] increasing.
Will the equilibrium shift, or will it not, when I crush a solid, and most importantly, why?
Thanks for reading and thanks in advance.
So I have been learning about equilibrium recently in class... and suppose I have the following reaction:
A(s) + B(aq) <-----> C(aq) + D(g)
[Note: in parentheses, I indicate the state of my substances]
So I was told recently that pulverizing my solid "A" would not change the position of equilibrium as nothing will happen... but I recalled our unit on Kinetics which says that increasing the surface area of my reactants will increase frequency of collisions which will increase the rate of the chemical reaction.
So, my impression is that pulverizing A would increase the surface area and increase the collisions for the forward reaction. Thus, the forward reaction would speed up, and a shift to the products would occur with [C] and [D] increasing.
Will the equilibrium shift, or will it not, when I crush a solid, and most importantly, why?
Thanks for reading and thanks in advance.