A certain substance X decomposes. It is found that 50% of X remains after 100 minutes. How much X remains after 200 minutes if the reaction order with respect to X is a) zero order b) first order c) second order

Zero Order: t (1/2) = [A]o / (2*k), so k = [A]o / (2 * t (1/2)), where t (1/2) = 100 min. and [A]o=100% So incorporate condition. A = -.0005 x 200 mins+1 of introductory fixation approaches zero percent.

Initially Order: Half-life condition k=.00693 remaining with 100 mins half life

Coordinate condition lnx = -.00693x200mins + ln 1 squares with lnx = - 1.386. Take to the e to counteract ln so x=.25 or 25%

Second Order Initial Concentration of 1, half-existence of 100 mins. K =.01. Coordinate condition 1/x =.01 x 200 + 1. X = 33.3%