Module 5: Rates and Equilibrium

Question 1

Give the 3 equations for calculating rate.

Answer 1

RATE = change in conc / time
RATE = 1 / time
RATE = y / x

Question 2

Give 4 properties we can use that mirrors concentration that we can measure experimentally.

Answer 2

1. pH meter
2. how quickly a gas is produced (gas syringe)
3. loss of mass as gas is produced
4. colour (colorimeter)

Question 3

Name and describe the orders of a reactant and what effect they have on the rate of reaction.

Answer 3

0 order: No influence on the rate of reaction 1st order: Proportional reaction (as the conc doubles, the rate also doubles)
2nd order: Squared relationship (double the conc, rate goes 4x faster)

Question 4

Give the basic layout of a rate reaction.

Answer 4

Rate = K [H] [Br] [O2]

Question 5

Describe what the following rate-concentration graphs will look like:

1. reaction is 0 order.
2. reaction is 1st order.
3. reaction is 2nd order.

Answer 5

0 order= straight horizontal line, no effect on rate
1st order= line of x=y diagonally across the graph
2nd order= curved line, starting at (0,0) and increasing gradually.

Question 6

What are the units for:

1. rate
2. [A]

Answer 6

RATE = moldm^-3s^-1 
[A] = moldm^-3

Question 7

Describe what the following conc-time graphs will look like:
1. reaction is 0 order
2. reaction is 1st order
3. reaction is 2nd order
how can you distinguish between 1st and 2nd order graphs?

Answer 7

0 order= slight negative gradient line
1st order= curved line decreasing from top left to bottom right.
2nd order= curved line decreasing from top left to bottom right.
1st order has a constant half life (within experimental error) and 2nd order does not have a constant half life.

Question 8

Define half life.

Answer 8

Half life is the time taken for the concentration to go down by half.

Question 9

Give the equation for K.

Answer 9

K= ln2 / t x 1/2

Question 10

Describe the Rate Determining Step.

Answer 10

The slowest step in the reaction that dictates how quickly the reaction can proceed.

Question 11

Give 2 reasons a reaction mechanism is unlikely to proceed in 1 single step.

Answer 11

1. Unlikely that a large number of particles will collide at the same time.
2. The stoichiometry of the equation does not match the rate equation.

Question 12

What is the effect of temperature on the rate constant, K?

Answer 12

Increase in temperature gives more energy to molecules, so collisions are more frequents and more collisions exceed the Ea of the reaction so are successful.
Raising the temperature speeds up the rate of reactions by increasing the rate constant, K.
Generally, doubling the rate will double the value of the rate constant, K.

Question 13

How can you calculate overall order of a reaction?

Answer 13

Overall order of a reaction is the sum of the individual orders.

E.G: rate=K[B] [C]^2.
Overall order is 1 + 2 = 3

Question 14

Define mole fraction and give the equation.

Answer 14

The amount of a given component within a reaction mixture.
Mole fraction is given by the symbol X.
Mole fraction X(A)=

no. of mols of substance A                      ----------------------------------------------- total no of mols of all substances

Question 15

Define partial pressures and give the equation.

Answer 15

Amount of pressure exerted by an individual species within a reaction mixture.

Partial Pressure = Mole fraction of x P (total pressure)
of componnet A component A

Question 16

What does it suggest if K is greater than 1?

Answer 16

The reaction is product-favoured.

Equilibrium lies on the right hand side.

Question 17

What does it suggest if K is less than 1?

Answer 17

The reaction is reactant-favoured.

Equilibrium lies on the left hand side.

Question 18

How do changes in temperature affect K?

Answer 18

• -> If the temperature changes, the value of K changes.
• -> If K increases as temperature increases, the position of equilibrium moves to the right hand side and the forward reaction is endothermic.
• -> If K decreases as temperature increases, the position of equilibrium shifts to the left hand side and the forward reaction is exothermic.

Question 19

How do changes in concentration and pressure affect K?

Answer 19

Value of K is unaffected by changes in concentration and pressure.

Question 20

Take the equation N2O4 (g) —> 2NO2 (g)
(it is in equilibrium)
What happens if the concentration of N2O4 is doubled?

Answer 20

• -> System is no longer in equilibrium
• -> Equilibrium position must shift to restore the original value of K
• -> It must increase [NO2] and decrease [N2O4]
• -> Causes a shift in the equilibrium position from left to right.

Question 21

Take the equation N2O4 (g) --> 2NO2 (g)
(it is in equilibrium)
Mols of N2O4 = 0.4
Mols of NO2 = 0.3
What happens if pressure of both components is doubled?

Answer 21

K= [NO2]^2 = (0.3^2) x2 = (0.6^2)
————- ———– ———-
[N2O4] (0.4) (0.8)

–> System is no longer in equilibrium.
–> Equilibrium must shift to restore the original value of K
–> It must decrease [NO2] (on top) and increase [N2O4] (on bottom)
–> Causes a shift in equilibrium position from right to left.
This is consistent with Le Chatelier’s as it moves in the direction of fewer mols.

Question 22

How does the presence of a catalyst affect K?

Answer 22

Catalysts have no affect on K.

They only affect the time taken for equilibrium to be established, not the position of equilibrium.