Module 3 Rates of Chemical Reactions

 

1. What is the rate of a reaction?

 

Solution:

The rate of a reaction is the amount of reactant that disappears or product that appears per unit time. Mathematically it is the ratio of the change in the amount of a substance to the change in time.

 

 

Example 1 Mg(s) + 2 HCl(aq) MgCl2(aq) + H2

 

 

a. A student places a piece of magnesium in acid and measures 0.12 moles of hydrogen gas after 2.0 minutes. Six minutes after having added the Mg, a total of 0.36 moles of hydrogen were collected. Find the average rate of production of hydrogen gas between 2.0 and 6.0 minutes.

 

Solution:

Rate = DH2/ Dt = [0.36- 0.12] / [6.0-2.0] = 0.060 moles of H2/minute

 

 

b. Find the average rate of disappearance of hydrochloric acid between 2.0 and 6.0 minutes.

 

Solution:

If we examine the ratio of HCl/ H2 in the balanced equation we notice that it is 2/1. So the rate of disappearance of HCl will be twice the rate of hydrogen production.

 

Answer: -0.060 moles of H2/minute [2 moles of HCl/ mole of H2] = -0.12 mole HCl/minute. The negative indicates that the reactant is being used up.

 

 

Example 2 A student places 5.0 moles of hydrogen and 5.0 moles of iodine into a 2.0 L container and measures the amount of HI formed according to:

 

H2(g) + I2(g) 2 HI (g)

 

 

Heres her data:

Time (minutes)

0.0

1.0

2.0

3.0

4.0

5.0

6.0

Moles of HI present

0.0

4.0

6.0

7.4

8.4

9.0

9.2

 

 

Graph the amount of leftover H2 versus time (for the first 6.0 minutes)and find the average rate of hydrogen consumed.

 

Solution:

Time (minutes)

0.0

1.0

2.0

3.0

4.0

5.0

6.0

Moles of HI present

0.0

4.0

6.0

7.4

8.4

9.0

9.2

Moles of hydrogen that reacted

 

 

 

 

=HI/2=

0

=HI/2=

4.0/2.0=2.0

3.0

3.7

4.2

4.5

4.6

Moles of hydrogen left over

 

 

Initially there were 5.0 moles

(see question), so now we have 5.0 0 = 5.0 moles

5.0-2.0=3.0

2.0

1.3

0.8

0.5

0.4

 

We use the third row to obtain the rate:

Rate = [4.6-0]/[6.0-0]=0.77 moles of hydrogen/minute.

 

But to graph we need the last row of data since we are being asked for leftover hydrogen.

 

Notice that since the points on the graph generate a curve, the rate at which hydrogen disappears keeps changing. That's why we refer to the rate we calculated as an average rate. To get the instantaneous rate (the rate at any given instant) we would have to draw a tangent line to the curve at that given moment, and obtain the slope of the tangent line. If you imagine successive tangent lines drawn to the curve from left to right, the slopes of the tangents become progressively more gentle. In other words, the rate at which hydrogen disappears keeps decreasing with time.