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: Start from the rate law. For a second order reaction \(\text{rate}=k[A]^2\), so \(k=\dfrac{\text{rate}}{[A]^2}\).
Step 2: Plug in units. Rate has units \(mol\,L^{-1}\,s^{-1}\) and \([A]^2\) has units \((mol\,L^{-1})^2 = mol^2\,L^{-2}\).
Step 3: Divide. \(k=\dfrac{mol\,L^{-1}\,s^{-1}}{mol^2\,L^{-2}} = mol^{-1}\,L\,s^{-1}\).
Step 4: Conclusion. In words this is \(mole^{-1}\,litre\,second^{-1}\), i.e. option (iv).
\[\boxed{mol^{-1}\,L\,s^{-1}}\]