Question

A graph between log t_1/2and log a (abscissa) 'a' being the initial conc. of A in the reaction.

A ®product, the rate law is –

• A. –$\frac{d\lbrack A\rbrack}{dt}$ = k[A]⁰
• B. –$\frac{d\lbrack A\rbrack}{dt}$ = k[A]
• C. –$\frac{d\lbrack A\rbrack}{dt}$ = k[A]²
• D. None of these

Answer 1

Answer: (C)

–$\frac{d\lbrack A\rbrack}{dt}$ = k$A$²

Answer 2

t_1/2 = k a^(1–n)

log t_1/2 = log k + (1 – n) log a

\ 1 – n = 1

n = 2

\ second order reaction