Following is the graph between log t 1 2 and log a ( a = initial concentration of reactant) for a given reaction at 27 o C . Hence, order of the reaction is (Here, t 1 / 2 is half-life)

Step-by-step solution

t 1 2 ∝ 1 a n − 1 for nth order reaction. t 1 2 = k a 1 − n log t 1 2 = 1 − n log a + log k Thus, graph between log t 1 2 and log a is linear slope. 1 − n = tan 45 o n = 0