Arrange the following in the increasing order of rate of the second order reaction
Reactants → products
(A) Reaction at 298 K
(B) Reaction at 298 K with catalyst
(C) Reaction at 500 K with catalyst and double the initial concentration of reactant
(D) Reaction at 500 K with double the initial concentration of reactant
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B), (D), (C)
For a second order reaction the rate law is:
\( \displaystyle \text{Rate} = k[\mathrm{R}]^{2} \)
So the rate can be raised in two ways. A larger rate constant k: a catalyst lowers the activation energy, and a higher temperature gives more molecules energy above the barrier; for many reactions k nearly doubles for every 10 K rise. A higher concentration: doubling [R] makes the rate \( 2^2 = 4 \) times.
(A) 298 K, no catalyst, initial concentration: the slowest.
(B) 298 K with a catalyst: k is larger, so faster than (A).
(D) 500 K and double concentration: a rise of about 200 K multiplies k many times over, and doubling [R] gives another factor of 4, so faster than (B).
(C) 500 K, double concentration and a catalyst: everything in (D) plus the catalyst, so the fastest.
Shortcut: (A) has none of the three factors and must be slowest; (C) has all three and must be fastest. Only Option (1) starts with (A) and ends with (C).