CBSE Class 12 · 2015 · Delhi · Chemical KineticsLong Answer

For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:

t (s) | 0 | 30 | 60
[CH

_{3}

COOCH

_{3}

] (mol L

^{- 1}

) | 0.60 | 0.30 | 0.15

i) Show that it follows a pseudo first order reaction, as the concentration of water remains constant.
ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds. (Given: log 2 = 0.3010, log 4 = 0.6021)

Answer: (i) k is constant (t₁/₂ = 30 s at every stage) → pseudo first order. (ii) Average rate = 5 × 10⁻³ mol L⁻¹ s⁻¹.

Solution

(i) Test the first order equation

k = \frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ]}

(water, in large excess, is constant and merges into k):
For

t = 30

k = \frac{2.303}{30} lo g \frac{0.60}{0.30} = \frac{2.303 \times 0.3010}{30} = 0.0231 s^{- 1}

For

t = 60

k = \frac{2.303}{60} lo g \frac{0.60}{0.15} = \frac{2.303 \times 0.6021}{60} = 0.0231 s^{- 1}

k is constant (equivalently, the half-life is a constant 30 s), so the reaction is pseudo first order.
(ii) Average rate between 30 and 60 s:

rate = - \frac{Δ [ R ]}{Δ t} = \frac{0.30 - 0.15}{60 - 30} = \frac{0.15}{30}

= 5 \times 1 0^{- 3} mol L^{- 1} s^{- 1}

CBSE Class 12 Board · 2015 · Delhi · Subjective

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CBSE Class 12 · 2015 · Delhi · Chemical KineticsLong Answer

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