{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

chemtest2 - rate = concentration time C=Cfinal-Cinitial...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
- rate = concentration/ time - C=C final -C initial - concentration units: mol/L (M) - rate units: mol/Lxtime - general rate expression: rate = aA + bB  cC + dD rate = -1/a [A]/ t = -1/b [B]/ t = 1/c [C]/ t = 1/d [D]/ t - iodine clock reaction H 2 O 2 (aq) + 2I - (aq) + 2H 3 O + (aq) 4H 2 O (l) + I 2 (aq) as soon as I 2 is formed, vitamin C rapidly reduces it to I - 2 H 2 O (l) + I 2 (aq) + C 6 H 8 O 6 (aq) C 6 H 6 O 6 (aq) + 2H 3 O + (aq) + 2I - (aq) - when solution is less concentrated, takes longer to change color - when solution is hotter, changes color more quickly - catalysts: accelerate reactions but are not consumed themselves - rate of rxn = k[A] m [B] n - if a homogeneous catalytic species is present, may be included in rate eq but not in balanced - determining initial rate: [products]/ time or - [reactants]/ time - [R] /[R] 0 = fraction remaining zero order first order second order [R] 0 –[R] = kt ln([R] /[R] 0 ) = -kt 1/[R] – 1/[R] = kt [R] = -kt + [R] 0 ln [R] = -kt + [R] 0 1/[R] = +kt + 1/[R] 0 y = mx + b y = mx + b y = mx + b ****table 15.1 pg 689**** - half life – time req for the concentration of a reactant to decrease ½ initial - longer half life, slower the rxn - first order: t 1/2 = 0.693/k - zero order: t 1/2 = [R] 0 /2k - second order: t 1/2 = 1/k[R] 0 - collision theory of rxn rates - reacting molecules must collide with one another
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern