Question

For a reaction type \(A + B \rightarrow\) products, it is observed that doubling concentration of \(A\) causes the reaction rate to be four times as great, but doubling amount of \(B\) does not affect the rate. The unit of rate constant is

• A. \(s^{-1}\)
• B. \(s^{-1}\,mol\,L^{-1}\)
• C. \(s^{-1}\,mol^{-1}\,L\)
• D. \(s^{-1}\,mol^{-2}\,L^2\)

Hint:

If doubling concentration causes rate to become 4 times, order is 2. If doubling has no effect, order is 0. For overall order 2, unit of \(k\) is \(L\,mol^{-1}\,s^{-1}\).

Answer 1

The Correct Option is C

Step 1: Write rate law form.
\[ \text{Rate} = k[A]^m[B]^n \] Step 2: Find order with respect to \(A\).
Doubling \([A]\) makes rate four times:
\[ 2^m = 4 \Rightarrow m = 2 \] Step 3: Find order with respect to \(B\).
Doubling \(B\) does not change rate:
\[ 2^n = 1 \Rightarrow n = 0 \] Step 4: Total order.
\[ \text{Order} = m+n = 2+0 = 2 \] Step 5: Unit of rate constant for second order reaction.
Rate unit:
\[ mol\,L^{-1}\,s^{-1} \] For second order:
\[ k = \frac{\text{Rate}}{[A]^2} = \frac{mol\,L^{-1}\,s^{-1}}{(mol\,L^{-1})^2} = L\,mol^{-1}\,s^{-1} \] So unit is:
\[ s^{-1}\,mol^{-1}\,L \] Final Answer: \[ \boxed{s^{-1}\,mol^{-1}\,L} \]