Consider the reaction, A+B→products. When [A] is tripled keeping [B] constant the rate of the reaction becomes three times as compared to initial rate. When [B] is doubled keeping the [A] constant the rate of the reaction doubles as compared to initial rate. Rate law can be written as
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a
rate=K[A]13[B]1
b
rate=K[A]1[B]1
c
rate=K[A]-13[B]1
d
rate=K[A]-1[B]1
answer is B.
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Detailed Solution
order w.r.t ‘A’ rate=KCn; rate∝Cn; 3=3x; x=1;Order w.r.t. ‘B’rate∝Cn; 2=2y; y=1;rate=K[A]1 [B]1 Overall order = 1 + 1=two ;
Consider the reaction, A+B→products. When [A] is tripled keeping [B] constant the rate of the reaction becomes three times as compared to initial rate. When [B] is doubled keeping the [A] constant the rate of the reaction doubles as compared to initial rate. Rate law can be written as