JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

Analysis of the graph: The graph shows reactant concentration [R] vs time t for three reactions starting from the same initial concentration. Curve 1 is a straight line with negative slope. For a zero-order reaction, the integrated rate law is [R] = [R]_0 - kt, which is a straight line in [R] vs t. Hence, reaction 1 is a zero-order reaction. Curves 2 and 3 are non-linear (exponential-type decays), so they correspond to reactions of order greater than zero (e.g., first or second order). Evaluating the options: (1) The order of all three reactions is the same. Reaction 1 is zero order while reactions 2 and 3 are of higher order, so this statement is incorrect. (2) The rate constant of reaction 3 is larger than that of reaction 2, assuming both have the same order. Assuming both curves 2 and 3 correspond to the same order (say first order), the integrated rate law is [R] = [R]_0 e^ -kt . At any given time t, the curve for reaction 3 lies below that of reaction 2, i.e., [R]_3 k_2, and this statement is correct. (3) The SI unit of the rate constant of reaction 1 is s ^ -1 . For a zero-order reaction, Rate = k[R]^0 = k, so the unit of k is that of rate, i.e., mol L ^ -1 s ^ -1 . The unit s ^ -1 corresponds to a first-order reaction, so this statement is incorrect. (4) Thermal decomposition of HI on a gold surface is an example of reaction 2. The thermal decomposition of HI on a gold surface is a well-known zero-order reaction, which corresponds to curve 1, not curve 2. Hence this statement is incorrect. Therefore, the correct option is (2).