1. Chem 1310: Introduction to physical chemistry Part 2: exercise 9, p657 Kinetics involving solids

2. Dissolving Al in base A cubic piece of Al, edges of 1 cm, is put into a 9M solution of NaOH. It starts to react, evolving H 2. Assume the reaction occurs at the surface. What would happen to the reaction rate if we cut the cube into two equal parts? How can we get the highest possible reaction rate?

3. Dissolving Al in base Creating the balanced reaction equation: □Al + □OH - + □H 2 O  □Al(OH) 4 - + □H 2 1Al + □OH - + □H 2 O  1Al(OH) 4 - + □H 2 1Al + 1OH - + □H 2 O  1Al(OH) 4 - + □H 2 1Al + 1OH - + 3H 2 O  1Al(OH) 4 - + □H 2 1Al + 1OH - + 3H 2 O  1Al(OH) 4 - + 1½H 2 2Al + 2OH - + 6H 2 O  2Al(OH) 4 - + 3H 2

4. Dissolving Al in base Reaction happens at surface. Rate should be proportional to surface area. Surface of cube = 6 cm 2 (6 faces, each 1*1 cm). Cut cube in two: creates two new faces of 1 cm 2 each. New surface area = 8 cm 2. Expected new rate: 8/6 = 1.33 times old rate.

5. Dissolving Al in base How to get the highest rate (without heating)? Cut into tiny pieces: same mass, much more surface area  much higher rate. Stirring and increasing the concentration of NaOH might also help.