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Q.i) Mention two differences between order and molecularity of chemical reaction. ii) 2 N2O5(g) -> 4 NO2(g) + O2(g). For this reaction the concentration of NO2 increases by 4×10^-2 mol/L in 5 seconds. Calculate the rate of reaction and rate of disappearance of N2O5. iii) If the unit of rate and rate constant of a chemical reaction is same then what will be the order of the reaction?

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2026Subjective· 5mImportance★★★★★
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Order is experimental (any value, from the rate law); molecularity is the integer count of colliding species in an elementary step. From d[NO2]/dt = 8x10^-3, the reaction rate is 2x10^-3 and -d[N2O5]/dt = 4x10^-3 mol L^-1 s^-1; equal units of rate and k mean zero order.

(i) Order vs molecularity - two differences

  1. Order is determined experimentally as the sum of the powers of the concentration terms in the rate law; molecularity is a theoretical count of the reactant species that collide in a single elementary step.
  2. Order may be zero, fractional or integral, and applies to overall (even complex, multi-step) reactions; molecularity is always a whole number (1, 2, 3...), can never be zero or fractional, and is defined only for an elementary reaction.

(ii) Rate calculation for 2N2O5 -> 4NO2 + O2

Rate of formation of NO2 = d[NO2]/dt = (4x10^-2 mol L^-1) / (5 s) = 8x10^-3 mol L^-1 s^-1.

The unique rate of reaction uses the stoichiometric coefficients:

rate = -(1/2) d[N2O5]/dt = +(1/4) d[NO2]/dt = +d[O2]/dt

So rate = (1/4)(8x10^-3) = 2x10^-3 mol L^-1 s^-1.

Rate of disappearance of N2O5 = -d[N2O5]/dt = 2 x rate = 2 x 2x10^-3 = 4x10^-3 mol L^-1 s^-1.

(iii) Units of rate = units of rate constant

Rate always has units mol L^-1 s^-1. The rate constant k has units (mol L^-1)^(1-n) s^-1 for order n. These coincide only when 1 - n = 0, i.e. n = 0 (zero-order reaction), for which rate = k.

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