1.

For a reactionAtoB with activation energy E_(a) and rate constant k=Ae^(-Ea//RT). The rate of the reaction (Rate =k[A]) increases by increasing the temperature because

Answer»

ACTIVATION ENERGY decreases with increase in TEMPERATURE
The FACTOR -Ea/RT increases
less number of collision take place
the value of [A] increases

Solution :If temperature increases, then `((-Ea)/(RT))` increases


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