18 The volume is 1 liter so all molar concentrations are numerically equal to

# 18 the volume is 1 liter so all molar concentrations

This preview shows page 3 - 4 out of 4 pages.

18. The volume is 1 liter so all molar concentrations are numerically equal to the number of moles. The stoichiometry is 2:2:2 for the NO, the NH 3 and the H 2 O. Therefore the moles of NO reacted must be equal to moles of NH 3 and H 2 O produced. That number is given, amount of H 2 O at equilibrium=0.00294 moles. This is also same as [NH 3 ] eq . The moles [NO] eq = starting amount minus the amount reacted: 0.00345 - 0.00294 = 5.10 x 10 - 4 M. Calculate amount of H 2 reacted and subtract that value from the original amount to find the amount left at equilibrium: amt reacted = NO mol 2 H mol 5 ) NO mol 00294 . 0 ( 2 = 0.00735 moles H 2 amt remaining: 0.0175 moles - 0.00735 moles H 2 in the 1 Liter volume = 0.0105 M Summary of results: [NO ]= 5.10 x 10 - 4 M; [H 2 ]= 0.0105 M; [NH 3 ] = 0.00294 M 19. (a) I use lines 1 and 2 to determine rate dependence on Br 2 under conditions where acetone is constant. x 2 trial 2 x 1 trial 2 ) ] Br ([ ) ] Br ([ 2 rate 1 rate = x 4 5 050 . 0 100 . 0 10 x 14 . 1 10 x 70 . 5 = - - log 0.5 = x log (2.00) therefore x = - 1 reaction rate is dependent on [Br 2 ] - 1 (b) use lines 2 and 3 to determine rate dependence on acetone under conditions where Br 2 is constant. x 3 trial x 2 trial ) ] acetone ([ ) ] acetone ([ 3 rate 2 rate = y 4 4 400 . 0 300 . 0 10 x 97 . 2 10 x 14 . 1 = - - log (0.38383) = y log (0.75) therefore y = 3.3 reaction rate is [acetone] 3.3 rate law: rate = k [acetone] 3.3 [Br 2 ] - 1 (c) Numerical value is calculated from data on any one line. I choose line 1. rate = k [acetone] 3.3 [Br 2 ] - 1 (1) 5.70x10 - 5 = k[0.300] 3.3 [0.100] - 1 k = 3.15 x 10 - 4 The other lines give the same answer to three sig figs.

#### You've reached the end of your free preview.

Want to read all 4 pages?

• Spring '08
• LEMASTER
• Chemistry, Mole, Chemical reaction, Moles

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern