Question

(b) Show that time required for completion 3/4th of a
first order reating is twice the time required for completion
of 1/2 of the reaction.​

Answer 1

Answer:

Since it is first order, thus

t= 2.303klog aa-xalso x =3a/4so the equation is t3/4= 2.303klog aa-3a/4.....1for x=1a/2t1/2= 2.303klog aa-a/2......2divide the 1 by 2, we gett3/4t1/2= 2.303klog aa-3a/42.303klog aa-a/2thus we gett3/4t1/2= log4log2t3/4t1/2= 2log2log2t3/4t1/2= 2 t3/4= 2t1/2

Answer 2

Answer:

Since it is first order, thus

t= 2.303klog aa-xalso x =3a/4so the equation is t3/4= 2.303klog aa-3a/4.....1for x=1a/2t1/2= 2.303klog aa-a/2......2divide the 1 by 2, we gett3/4t1/2= 2.303klog aa-3a/42.303klog aa-a/2thus we gett3/4t1/2= log4log2t3/4t1/2= 2log2log2t3/4t1/2= 2 t3/4= 2t1/2