The following graph is experimentally obtained for the reaction : A → 2B, at 25°C. The correct statement(s) for the reaction is/are : (Given : ln2 = 0.7)

Engineering

Chemistry

Integrated Rate Law

Question

The following graph is experimentally obtained for the reaction : A → 2B, at 25°C.

The correct statement(s) for the reaction is/are : (Given : ln2 = 0.7)

The initial concentration of 'A' was 'e' M.

Time for 87.5% reaction of 'A' is 21 min.

The time at which the concentration of 'A' and 'B' becomes equal is 7 min.

The rate of appearance of 'B' is $+ \frac{d [B]}{dt}$ = (0.2 min–1) [A].

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$\ln \frac{[A]}{M}$
ln [A]₀ – ln [A]_t = Kt
ln [A]_t = – Kt + ln [A]₀
– K = $\frac{- 1 .4 + 0 .7}{24 - 17} = - \frac{0 .7}{7} = - 0 .1$
K = 0.1
t_1/2 = $\frac{\ln 2}{K} = \frac{0 .7}{0 .1} = 7$
ROR = $\frac{- d [A]}{dt} = \frac{1}{2} \frac{d [B]}{dt}$ = K [A]
$\frac{d [B]}{dt} =$ 2K [A] = 0.2 [A]
ln [A]t = – Kt + ln [A]₀
0 = – 0.1 × 10 + ln [A]₀
ln [A]₀ = 1
[A]₀ = e

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