CBSE Class 12 · 2015 · Chemical KineticsLong Answer

a) For a reaction A + B → P, the rate is given by: Rate = k[A][B]

^{2}

.
i) How is the rate of reaction affected if the concentration of B is doubled?
ii) What is the overall order of reaction if A is present in large excess?
b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. (log 2 = 0.3010)

Answer: (a)(i) Rate ×4 (ii) Order = 2 (pseudo second order). (b) ≈ 100 min.

Solution

(a)(i) Rate

\propto [B]^{2}

; doubling [B] makes the rate

2^{2} = 4

times.
(ii) With A in large excess, [A] is effectively constant: rate

= k^{'} [B]^{2}

—

overall order = 2

(b) From the half-life:

k = \frac{0.693}{30} = 0.0231 mi n^{- 1}

For 90% completion (

[R] = [R]_{0} /10

t = \frac{2.303}{0.0231} lo g 10 = \frac{2.303}{0.0231}

t \approx 99.7 min (\approx 100 min)

CBSE Class 12 Board · 2015 · Subjective

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CBSE Class 12 · 2015 · Chemical KineticsLong Answer

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