For the reaction,Products, it is observed that.
Rate = k[A2[B]
Rate = k[A][B]2
Rate = k[A]2[B]2
Rate = k[A][B]
The rate constant k1 and k2 for two different reactions are 1016.e-2000/T and 1015.e-1000/T, respectively. The temperature at which k1 =k2 is
1000 k
2000 k
A chemical reaction is catalysed by a catalyst X.Hence,X.
Reduces enthalpy of the reaction.
Decreases rate constant of the reaction.
Increases activation energy of the reaction.
Does not affect equilibrium constant of the reaction.
1.732
3
1.02 x 10-4
3.4 x 105
The activation energy for a simple chemical reaction is Ea in forward direction. The activation energy for reverse reaction.
Can be less than or more than Ea.
Is always double of Ea.
Is negative of Ea.
Is always less than Ea.
Select the rate law that corresponds to data shown for the following reaction.
rate =k[B]3
rate =k[B]4
rate =k[A][B]3
rate =k[A]2[B]2
When a biochemical reaction is carried out in laboratory from outside of human body in the absence of enzyme, the rate of reaction obtained is 10-6 times,then activation energy of the reaction in the presence of enzyme is.
P is required.
Different from Ea obtained in laboratory
Cannot say any things
The reaction follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B 1 h. What is the time taken for the conversion of 0.9 mole of A to 0.675 mole of B ?
0.25 h
2 h
1 h
0.5 h