Solutions to Rate Constants

 

1. a. Since the rate is proportional to [I-1], it will also be half of the original rate.

b. If the rate remains constant then the ratio of the two rates = 1:

So 1 = 3x

x = 1/3.

So we need only 1/3 of the original amount of iodide.

 

2. a. If you quadruple the concentration of each reactant, then the

rate will become (4) (4)0.5 = 8 times the original rate. So the reaction will only take (1/8) as long.

2h =120 minutes

120 minutes/8 = 15 minutes.

 

b. 3 = (2)(x)0.5.

1.5 = (x)0.5.

2.25 = x (we just squared both sides)

So the molarity(concentration) of Br2 was made 2.25 times bigger. ( It's important to write this out; otherwise it may sound like the concentration of bromine was 2.25 moles/L. It's not, unless the original was 1 M.)

 

3. The overall reaction often conceals what really happened. There may be two or more steps leading to the reaction, of which only one is rate-determining. That will lead to an unexpected expression, as seen in the example with NO2 and CO. With enzymes, increasing the concentration of either reactant often has no impact once the few enzyme molecules are occupied. This is because the other extra molecules you've introduced would have to "wait for their turn" anyway.

 

4. 6.35 g/[63.59(g/mole)] = 0.10 moles of Cu reacted in 100 s. (60 + 40)

rate at which Cu disappears = 0.10/100 = 0.0010 moles/s.

 

5. The question was:

 

While studying the rate of various chemical reactions, a student measured the rate at which certain metals react with different acids. One of the experiments involved combining a strip of solid magnesium, Mg(s), with a hydrochloric acid solution, HCl(aq). The student made the following observations:

 

- Mass of the magnesium strip used 1.78 10-2 g

- Atmospheric pressure in the room 101.3 kPa

- Room temperature 25.0C

- Temperature of the acidic solution 25.0C

- Duration of the reaction 6 min 40 s

This chemical reaction is represented by the following equation:

 

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

  Under these conditions, what is the average rate of production of H2(g) in ml/min?

 

Solution:

 

0.0178 g / [ 24 (g/mole)] = 7.5 X 10-4 moles Mg

The ratio is 1:1, so we also have 7.5 X 10-4 moles of H2.

This formed in (6)(60) + 40 = 400 seconds.

So the rate at which hydrogen formed is 7.5 X 10-4 moles/400 s = 1.9 X 10-6 moles/s. Now use PV = nRT to convert moles to ml;

V= nRT/P = 1.9 X10-6(8.31)(25 + 273)/101.3 =0.00004644 L

Rate = 0.04 ml/s =