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Q.(a) Define order of reaction. How does order of a reaction differ from molecularity for a complex reaction ?

(b) A first order reaction is 50% complete in 25 minutes. Calculate the time for 80% completion of the reaction.
(OR)
(a) The decomposition of a hydrocarbon has value of rate constant as 2.5×104 s−12.5 \times 10^4\ s^{-1} at 27∘^{\circ}C. At what temperature would rate constant be 7.5×104 s−17.5 \times 10^4\ s^{-1} if energy of activation is 19.147×103 J mol−119.147 \times 10^3\ J\ mol^{-1} ?
(b) Write a condition under which a bimolecular reaction is kinetically first order. Give an example of such a reaction. (Given : log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990)
CBSECBSE Class XII Board 2019Subjective· 5mImportance★★★★★
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Part (a): order = experimental exponent sum (can be 0/fraction/integer) and, for a complex reaction, is set by the slow step; molecularity is the integer number of species in an elementary step. First-order 80% completion takes ≈58\approx58 min. Part (b): Arrhenius gives T2≈350T_2\approx350 K (77 ∘77\,^\circC); a bimolecular reaction is kinetically first order when one reactant is in large excess (e.g. ester hydrolysis).

Part (a)

(a) Order vs molecularity

Order of reaction: the sum of the exponents of the concentration terms in the experimental rate law. For rate =k[A]p[B]q=k[A]^p[B]^q, order =p+q=p+q; it may be zero, fractional or integral and can depend on conditions.

Molecularity: the number of reacting species that come together in a single elementary step; always a small positive integer (1, 2, rarely 3) and derived from the mechanism.

FeatureOrderMolecularity
Basisexperimental rate lawelementary step of the mechanism
Values0, fraction, integer1, 2, 3 only
Applies tooverall reactioneach elementary step

For a complex reaction, the overall order equals that of the rate-determining (slowest) step, which may differ from the molecularity of any individual step. So order and molecularity coincide only for genuinely elementary reactions.

(b) Time for 80% completion (first order)

From the half-life, k=0.693t1/2=0.69325=0.02772 min−1k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{25}=0.02772\ \text{min}^{-1}.

For 80% completion, 20% remains, so [A]0[A]=10020=5\dfrac{[A]_0}{[A]}=\dfrac{100}{20}=5:

t=2.303klog⁡5=2.303×0.69900.02772≈58.0 min.t=\frac{2.303}{k}\log 5=\frac{2.303\times0.6990}{0.02772}\approx 58.0\ \text{min}. …

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