The order of a reaction whose rate constant k = 0.693 x 10-6 L mol-1…

The correct answer is Option (3) → 2

The unit of the rate constant shows the order of a reaction. The rate always has the unit mol L⁻¹ s⁻¹. For a reaction of order n, Rate = k[A]ⁿ, so:

    \( \displaystyle k=\frac{\text{Rate}}{[A]^n} \)

    \( \displaystyle \text{unit of }k=\frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{(\mathrm{mol\,L^{-1}})^n}=(\mathrm{mol\,L^{-1}})^{1-n}\,\mathrm{s^{-1}} \)

• Zero order (n = 0): mol L⁻¹ s⁻¹

• First order (n = 1): s⁻¹

• Second order (n = 2): (mol L⁻¹)⁻¹ s⁻¹ = L mol⁻¹ s⁻¹

• Third order (n = 3): L² mol⁻² s⁻¹

The given k has the unit L mol⁻¹ s⁻¹ = (mol L⁻¹)⁻¹ s⁻¹. Comparing the powers of (mol L⁻¹):

    \( \displaystyle 1-n=-1 \)
    \( \displaystyle n=2 \)

So the reaction is of second order. The number 0.693 × 10⁻⁶ plays no part; only the unit decides the order.

Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Units of rate constant