For the reaction2N2O5(g) → 4NO2(g) + O2(g)The rate is expressed as:

The correct answer is Option (1) → \( \displaystyle -\frac{\Delta[\ce{N2O5}]}{2\,\Delta t}=\frac{\Delta[\ce{NO2}]}{4\,\Delta t}=\frac{\Delta[\ce{O2}]}{\Delta t} \)

The rate of a reaction must have one value, whichever substance is used to follow it. For a general reaction

    \( \displaystyle \ce{aA + bB -> cC + dD} \)

    \( \displaystyle \text{Rate}=-\frac{1}{a}\frac{\Delta[\ce{A}]}{\Delta t}=-\frac{1}{b}\frac{\Delta[\ce{B}]}{\Delta t}=+\frac{1}{c}\frac{\Delta[\ce{C}]}{\Delta t}=+\frac{1}{d}\frac{\Delta[\ce{D}]}{\Delta t} \)

For 2N₂O₅(g) → 4NO₂(g) + O₂(g), the coefficients are 2, 4 and 1:

    \( \displaystyle \text{Rate}=-\frac{\Delta[\ce{N2O5}]}{2\,\Delta t}=\frac{\Delta[\ce{NO2}]}{4\,\Delta t}=\frac{\Delta[\ce{O2}]}{\Delta t} \)

So the correct expression is Option (1).

Shortcut: Only the reactant \( \ce{N2O5} \) may carry the minus sign. Options (2), (3) and (4) all put the minus sign on \( \ce{O2} \), a product, so only Option (1) remains.

Why the other options are incorrect

• Option (2) \( \displaystyle \frac{2\Delta[\ce{N2O5}]}{\Delta t}=\frac{4\Delta[\ce{NO2}]}{\Delta t}=-\frac{\Delta[\ce{O2}]}{\Delta t} \) — Incorrect. The changes are multiplied by the coefficients instead of divided; the \( \ce{N2O5} \) term has no minus sign and the product \( \ce{O2} \) wrongly has one.

• Option (3) \( \displaystyle \frac{\Delta[\ce{N2O5}]}{2\,\Delta t}=\frac{\Delta[\ce{NO2}]}{4\,\Delta t}=-\frac{\Delta[\ce{O2}]}{\Delta t} \) — Incorrect. The coefficients are used correctly, but the \( \ce{N2O5} \) term has lost its minus sign and \( \ce{O2} \) has wrongly received one.

• Option (4) \( \displaystyle \frac{\Delta[\ce{N2O5}]}{\Delta t}=\frac{\Delta[\ce{NO2}]}{\Delta t}=-\frac{\Delta[\ce{O2}]}{\Delta t} \) — Incorrect. The coefficients are ignored and the signs are wrong. \( \ce{N2O5} \) is used up twice as fast as \( \ce{O2} \) forms, so these terms cannot be equal.

NCERT (2026-27 edition):
Rate of a Chemical Reaction – …