For the first order reaction R → P, deduce the integrated form of the rate law. What will be the nature of the curve when concentration is plotted against time for such a reaction?

Solution

For the reaction

R → P

R a t e = - \frac{d [R]}{d t} = K [R] O r \frac{1}{[R]} d [R] = - k d t O n i n t e g r a t i n g b o t h s i d e s, w e g e t \int \frac{1 d [R]}{R} = - k \int d t I n [R] = - k t + I . . . . . (1) W h e r e I = i n t e g r a t i o n c o n s \tan t A t t = 0, [R] = [R]_{0} i n t i a l c o n c o f t h e r e a c \tan t p u r t t i n g t h i s i n e q (1) g i v e s I n [R]_{0} = - k x o + I a n d I = I n [R]_{0} s u b s t i t u t i n g t h e v a l u e o f I i n e q (1) g i v e s I n [R] = - K t + I n [R]_{0} C h a n g i n g i n t o \log 10 w e h a v e \log [R] = - \frac{k}{2.303} t + \log [R]_{0} W h e n \log [R] i s p l o t t e d a g a i n s t t i t g i v e s a s t r a i g h t l i n e w i t h s l o p e = \frac{- k}{2.303} a n d y i n t e r c e p t = \log [R]_{0} k = \frac{2.303}{t} \log \frac{[R]}{[R]_{0}}

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