For nth order reaction (dx/dt) = Rate = k[A0]^n. Graph between log(rate) against log[A0] is of the type. Lines P, Q, R, S are for the order:
P4,Q7,R6,S5
P4,Q5,R6,S7
P5,Q6,R7,S4
P7,Q6,R5,S52
Solution:
Given that, Rate = k[A0]^n — 1 Taking log on both sides, we get log rate = n log[A]o + log k As n increases the slope of graph increases. Hence, P7,Q6,R5,S52