- #1
nothing123
- 97
- 0
Hi,
So if we have a two-step mechanism that goes something like this:
1) A + B -> C
2) C -> D
-------------
A + B -> D
and the second step is the slow step, the reaction rate would be r = k2[C] correct? Well, seeing as reactants A and B are zero order, varying their concentrations wouldn't affect the rate. However, one of the questions I encountered said if you ADD A or B or DECREASE A or B, the rate subsequently increases and decreases respectively. I understand the reasoning for this since more A or B would drive the reaction to the right and thus increasing the concentration of intermediate C, which is important in determining the rate. But really, what's the difference between varying the concentration of A or B and adding A or B; if we add A, are we not effectively increasing its concentration?
I got to second question kind of related to this. We have a reaction A + B -> C with equilibrium constant: Kc = [C]/[A]. Now, if we add reactant A, BUT at the same concentration as the equilibrium concentration, then according to the equation, the reaction does not shift to the right, correct?
Thanks.
So if we have a two-step mechanism that goes something like this:
1) A + B -> C
2) C -> D
-------------
A + B -> D
and the second step is the slow step, the reaction rate would be r = k2[C] correct? Well, seeing as reactants A and B are zero order, varying their concentrations wouldn't affect the rate. However, one of the questions I encountered said if you ADD A or B or DECREASE A or B, the rate subsequently increases and decreases respectively. I understand the reasoning for this since more A or B would drive the reaction to the right and thus increasing the concentration of intermediate C, which is important in determining the rate. But really, what's the difference between varying the concentration of A or B and adding A or B; if we add A, are we not effectively increasing its concentration?
I got to second question kind of related to this. We have a reaction A + B -> C with equilibrium constant: Kc = [C]/[A]. Now, if we add reactant A, BUT at the same concentration as the equilibrium concentration, then according to the equation, the reaction does not shift to the right, correct?
Thanks.