See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

As we know, 1 n 1/2 0 t a       1 n 1/2 0 t c. a c constant       1/2 0 logt logc 1 n loga      o 1 n tan 45 1 n 1 n 0         Now, for K, As we know for zero order reaction, 0 1/2 a t 2K  0 1/2 0 a 1 logt log loga log 2K 2K           … (1) From graph, 1 log 2 2K  1 1 1 K moll sec 200    1 250 K 250