[T]

Hello, I have attached a photo of a graph. The reaction equation was:
PbI₂  Pb²⁺ + 2I^-

My question is, if the reverse reaction slows down at time t, why won't the forwards reaction also slow down? At t, the reverse reaction decreased, but the forwards reaction was still constant. This meant that shortly after t, the concentration of PbI₂ will decrease as a result because the forwards reaction>reverse reaction. If the concentration of PbI₂ decreased, won't the forwards reaction decrease too because of the collision and activation energy theory?

Thanks, sorry if I am confusing.

[mjc123]

You don't specify the states of the reagents, but I assume the reaction is
PbI₂(s)  Pb²⁺(aq) + 2I^-(aq)
If PbI₂ is solid, not in solution (and undissociated PbI₂ has no solubility, as far as I know) then it has no "concentration" to decrease. By convention the activity of all pure solid or liquid phases has a value of 1, however much or little is present. As it reacts, the activity does not change, so nor does the rate of reaction. Even if the equilibrium is shifted to the right by reducing the rate of the reverse reaction (how? not by lowering the temperature, because that would reduce the forward reaction rate too), the rate of the forward reaction remains unchanged.