A first-order reaction has k=4.0x10^-5 s^-1 at 300 K and 1.6x10^-3 s^-1 at 350 K. Using R=8.314 J mol^-1K^-1, ln 40=3.689 and ln10=2.303, calculate log_10(A/s^-1) to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3ck161

A first-order reaction has $k = 4.0 \times 1 0^{- 5} s^{- 1}$ at 300 K and $1.6 \times 1 0^{- 3} s^{- 1}$ at 350 K. Using $R = 8.314 J mo l^{- 1} K^{- 1}$ , $ln 40 = 3.689$ and $ln 10 = 2.303$ , calculate $lo g_{10} (A / s^{- 1})$ to two decimal places.

Numerical answer type, tolerance 0.04.

E_{a} = \frac{R l n 40}{1/300 - 1/350} = \frac{8.314 \times 3.689}{0.0004762} = 64406 J mo l^{- 1}

. Then

ln A = ln k_{1} + E_{a} / R T_{1} = ln (4.0 \times 1 0^{- 5}) + 25.822 = 15.696

, so

lo g_{10} A = 6.82