Question

For the reaction: \[ R \rightarrow P \] The variation of concentration of reactant \([R]\) with time is shown below. Statements: I. Curve 1 will be observed when order of reaction is 1. II. If curve 2 & curve 3 are observed for \(1^{st}\) order then rate constant for \(3^{rd}\) curve will be greater than rate constant for \(2^{nd}\) curve. III. Decomposition of HI on gold surface represent curve (1). Select the option representing correct set of statements: 

• A. I, II, III
• B. I, II
• C. II, III
• D. I, III

Answer 1

The Correct Option is C

Concept: The concentration–time graph differs for different orders of reactions. • Zero order: \([R]\) decreases linearly with time.
• First order: exponential decay curve.
• Higher rate constant gives faster decrease of concentration.
Step 1: Analyse curve 1 Curve 1 is a straight line decrease of concentration with time. This corresponds to zero order reaction, not first order. Hence statement I is incorrect.
Step 2: Compare curve 2 and curve 3 For first order reactions, steeper decay corresponds to larger rate constant. Curve 3 decreases faster than curve 2. Therefore, \[ k_3 > k_2 \] Statement II is correct.
Step 3: Example of zero order reaction Decomposition of HI on gold surface follows zero order kinetics. Hence it corresponds to curve 1. Statement III is correct. Thus, \[ \boxed{\text{Correct statements: II and III}} \]