Question

The velocity (rate) constant of a second order reaction is generally expressed as:

• A. mole litre second
• B. \(mole^{-1}\,litre^{-1}\,second^{-1}\)
• C. \(mole\,litre^{-1}\,second^{-1}\)
• D. \(mole^{-1}\,litre\,second^{-1}\)

Hint:

Second order: \(k = mol^{(1-n)}L^{(n-1)}s^{-1}\) with \(n=2\) gives \(mol^{-1}L\,s^{-1}\).

Answer 1

The Correct Option is D

Step 1: Formula for the unit of \(k\). For order \(n\): \[k = mol^{(1-n)}\,L^{(n-1)}\,s^{-1}.\]
Step 2: Substitute \(n=2\). \(k = mol^{(1-2)}\,L^{(2-1)}\,s^{-1} = mol^{-1}\,L\,s^{-1}\).
Step 3: Read as words. \(mol^{-1}\,L\,s^{-1} = mole^{-1}\,litre\,second^{-1}\).
Step 4: Match option. This is option (iv). Option (ii) \(mole^{-1}\,litre^{-1}\,second^{-1}\) is wrong (extra \(litre^{-1}\)); option (iii) is the zero-order unit; option (i) has no reciprocal terms at all.
\[\boxed{\text{mole}^{-1}\,\text{litre}\,\text{second}^{-1}\ (\text{option iv})}\]