The inversion of sucrose is pseudo-first-order with k=5.0x10^-3 min^-1. The initial rotation is +68.0 deg and the final rotation is -20.0 deg. Calculate the observed rotation (in degrees, to one decimal place) after 100 min. Use e^-0.5=0.6065.

JEE Main levelNumerical answerDifficulty 2 of 3ck129

The inversion of sucrose is pseudo-first-order with $k = 5.0 \times 1 0^{- 3} mi n^{- 1}$ . The initial rotation is $+ 68. 0^{\circ}$ and the final rotation is $- 20. 0^{\circ}$ . Calculate the observed rotation (in degrees, to one decimal place) after 100 min. Use $e^{- 0.5} = 0.6065$ .

Numerical answer type, answer in degrees, tolerance 0.3.

r_{t} - r_{\infty} = (r_{0} - r_{\infty}) e^{- k t} = 88.0 \times 0.6065 = 53.37

. So

r_{t} = - 20.0 + 53.37 = 33. 4^{\circ}