CHEMICAL KINETICS GRAPHICAL DIFFERENTIAL RATE LAWS GCHEM II LAB CONCEPT ANSWERS [A] versus Time straight line: 0th order slope = - k ln [A] versus Time straight line: 1 st order slope = - k 1/[A] versus Time straight line: 2 nd order slope = +k 1) A straight line graph developed when ln of concentration versus time was plotted for the decomposition of 2NO 2 (g) à 2NO(g) + 2O2(g) at 436 K. If the rate constant was 2.34 /sec, what is the reaction rate when concentration of NO 2 was 1.45 M. Reaction Rate = k [NO 2 ] N ln [NO 2 ] versus time results straight line – therefore 1 st order reaction Reaction Rate = k [NO 2 ] 1 Reaction Rate = 2.34[1.45] 1 Reaction Rate = 3.393 3.39 M/sec 2) Consider decomposition of hydrogen peroxide: H 2 O 2 (l) à 2 H 2 O(l) + O 2 (g). When the ln [H 2 O 2 ] versus time was plotted, the result was a straight line. What is the rate law and what is rate constant for the decomposition if the reaction rate is 1.2 x 10 –3 M/sec and concentration of H 2 O 2 is .75 M
This is the end of the preview. Sign up
access the rest of the document.