For A + B -> P (rate = k[A][B], k = 0.50 L mol^-1 s^-1) the initial concentrations are [A]_0 = 0.30 mol L^-1 and [B]_0 = 0.20 mol L^-1. Using kt = (1)/(a-b)ln(b(a-x))/(a(b-x)) with 2.303, log 2 = 0.3010 and log 3 = 0.4771, find the time (in seconds, two decimals) for 0.10 mol L^-1 of B to react.

JEE Advanced levelNumerical answerDifficulty 3 of 3ck097

For $A + B P$ (rate $= k [A] [B]$ , $k = 0.50 L mo l^{- 1} s^{- 1}$ ) the initial concentrations are $[A]_{0} = 0.30 mol L^{- 1}$ and $[B]_{0} = 0.20 mol L^{- 1}$ . Using $k t = \frac{1}{a - b} ln \frac{b ( a - x )}{a ( b - x )}$ with $2.303$ , $lo g 2 = 0.3010$ and $lo g 3 = 0.4771$ , find the time (in seconds, two decimals) for $0.10 mol L^{- 1}$ of $B$ to react.

Numerical answer type, answer in s, tolerance 0.03.

Here

a - b = 0.10

and

\frac{b ( a - x )}{a ( b - x )} = \frac{0.20 \times 0.20}{0.30 \times 0.10} = \frac{4}{3}

. Then

ln (4/3) = 2.303 (2 \times 0.3010 - 0.4771) = 2.303 \times 0.1249 = 0.2876

, and

t = \frac{1}{0.50 \times 0.10} \times 0.2876 = 5.75 s