00:01
Hello students, let's discuss the question here.
00:03
In this question, it is given that we have an equilibrium that is s to p.
00:10
For forward reaction, equilibrium constant value is k1 and for reverse reaction this is k2.
00:16
Now we have to observe that work will be the effect on general reaction if we add enzyme in it.
00:23
We know that enzyme, this will work as catalyst in this case.
00:30
Due to this, what will happen that activation energy of the system, this will be lowered? this is due to the fact that the work will be done by enzyme itself.
00:46
That is why the activation energy will lower due to the addition of enzyme in this case.
00:54
Next is the formation of transition state.
00:58
It will be promoted.
00:59
This is again due to the catalytic features of enzyme.
01:03
In this case and due to this the value of k2 that is rate constant of the reverse reaction this will increase in this situation hence from the given options the option that are correct that is the formation of transition state is promoted the activation energy of the reaction is lowered and the rate constant for the reverse reaction is increased while the option that says the concentration of product is increased, that is not true because enzyme it helps to attain the product very soon, that is by loading the activation energy, it won't change the concentration of the product, it will just change the rate of the reaction...