Answer to Question #259598 in General Chemistry for yel

Question

A cube of aluminum 1.0cm on each edge is plaved into 9-M NaOH(aq), and the rate at which H2 gas is given off is measured.

a) By what factors will this reaction rate change if the aluminum cube is cut exactly in half and the two halves are placed in the solution? Assume that the reaction rate is proportional to the surface area, and that all of the surface of the aluminum is in contact with the NaOH(aq).

b) If you had to speed up this reaction as much as you could without raising the temperature, what would you do to the aluminum?

Answer 1

Initial surface area= $(1×1)6$ =$6cm^2$

Area after cutting =$2(1×1)+4(1×0.5)$

= $8cm^2$

Area increases by factor $\frac {8}{6}=1.5$

$\therefore$Reaction rate increases by same factor =$1.5$

Precipitation of $Al(OH)_3$ crystals on the aluminium surface affect the rate of reaction.

When this is reduced by agitation,the reaction rate increases

The second way is reducing the thickness of Aluminium plates, therefore increasing the area of contact with $NaOH_{(aq)}$