For the reaction A→k2sec−1k1sec−1Bfollowing graph is givenk1=4×10−2sec−1which is/are correct statement(s) (ln 2=0.7, ln 8/7=0.14)

equilibrium constant , K = [ B ] [ A ] = 1 . 6 0 . 4 = 4 K = K f K b ⇒ K b = 4 × 10 - 2 4 = 10 - 2 s - 1 t 1 / 2 = 1 K 1 + K 2 ln a a - x = 1 5 × 10 - 2 ln 2 = 0 . 7 5 × 10 - 2 = 14 sec . t 10 % = 1 5 × 10 - 2 ln 1 . 6 1 . 6 - 0 . 2 = 1 5 × 10 - 2 ln 8 7 = 0 . 14 5 × 10 - 2 = 2 . 8 sec .

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For the reaction $A \to_{k_{2} \sec^{- 1}}^{k_{1} \sec^{- 1}} B$ following graph is given $k_{1} = 4 \times 10^{- 2} \sec^{- 1}$ which is/are correct statement(s) (ln 2=0.7, ln 8/7=0.14)

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Rate constant of backward reaction is $10^{- 2} \sec^{- 1}$

Time taken for the completion of 10% of initial conc. of A is 2.8 sec

Equilibrium constant is 4.0

Time taken for the completion of 50% of equilibrium conc. of B is 14 sec

answer is A, B, C, D.

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Detailed Solution

$equilibrium constant, K = \frac{[B]}{[A]} = \frac{1.6}{0.4} = 4 K = \frac{K_{f}}{K_{b}} \Rightarrow K_{b} = \frac{4 \times 10^{- 2}}{4} = 10^{- 2} s^{- 1} t_{1 / 2} = \frac{1}{K_{1} + K_{2}} \ln \frac{a}{a - x} = \frac{1}{5 \times 10^{- 2}} \ln 2 = \frac{0.7}{5 \times 10^{- 2}} = 14 \sec . t_{10 %} = \frac{1}{5 \times 10^{- 2}} \ln \frac{1.6}{1.6 - 0.2} = \frac{1}{5 \times 10^{- 2}} \ln \frac{8}{7} = \frac{0.14}{5 \times 10^{- 2}} = 2.8 \sec .$