2) Increasing the rate of the reverse reaction will cause a shift to the left.
Hopefully, If it is wrong my full apologies.
OR IF THATS WRONG, TRY THIS
a - Increasing the rate of the forward reaction will cause equilibrium to be more product-favored (i.e. shift to the right, not the left)
It’s hard which class are u
(missing part of your question):
when we have K = 1 x 10^-2 and [A] = 2 M & [B] = 3M & m= 2 & i = 1
So when the rate = K[A]^m [B]^i
and when we have m + i = 3 so the order of this reaction is 3 So the unit of K is L^2.mol^-2S^-1
So by substitution:
∴ the rate = (1x 10 ^-2 L^-2.mol^-2S^-1)*(2 mol.L^-1)^2*(3mol.L^-1)
= 0.12 mol.L^-1.S^-1
