IA on effect of concentration of NaOH on the rate of hydrogen production, between aluminium and sodium hydroxide.

1. net eqn
H2 fuel cell- acidic electrolyte
(-ve) (Anode) - Oxidation
2H2 → 4H+ + 4e−
+ ve (Cathode)- Reduction
4H+ + O2 + 4e− → 4H2O
2H2 + O2 → 2H2O O2
H2
PEM – made of Teflon
allow H+ ion to flow
Proton Exchange Membrane
H2O
Catalyst – platinum used anode/cathode
Effect of conc NaOH on the rate of hydrogen production
Hydrogen is used for fuel cell.
Aluminium is reactive – but will not react with acid as it forms a stable aluminium oxde layer.
NaOH is used to corrode away oxide layer, expose reactive AI metal.
NaOH acts as a catalyst. Source of H2 using aluminium.
Finding min conc, for rxn to happen.
Order of rxn will be determined.
Aluminium powder, strip, and foil will be tested
Al + 6H2O + 2NaOH → 2Na[AI(OH)4] + 3H2
2AI + 6H2O → 2AI(OH)3 + 3H2
Reaction mechanism
Procedure:
4ml 6M NaOH added to 0.01g AI foil
Conc NaOH – 2M, 3M, 4M, 5M, 6M
Pressure sensor will be used to measure the initial rate rxn.
Slope of pressure change over time – used as initial rate.
Assume conc NaOH in excess, doesnt change much over time.
AI foil

2. Effect of conc NaOH on the rate of hydrogen production
All rxn water bath to control temp at 23C Rxn happen at 2M NaOH
Pressure sensor to measure rate
Conc
NaOH/M
Rate
kPa/s-1
2 0.00504
3 0.00805
4 0.0106
5 0.0150
6 0.0313
Data collected.
Slope was taken for 1st 100s
y = 0.0016x1.5147
R² = 0.9148
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0 1 2 3 4 5 6 7
Rate
conc/M
Conc/M vs Rate/kPas-1
Outlier at 6M NaOH, rate too high due to exothermic rxn
Power fit taken, y= 0.0016x1.5
To find the order – Rate = k[NaOH]1.5
Order is 1.5

3. Effect of conc NaOH on the rate of hydrogen production
All rxn water bath to control temp at 23C Rxn happen at 2M NaOH
Pressure sensor to measure rate
Conc
NaOH/M
Rate
kPa/s-1
2 0.00504
3 0.00805
4 0.0106
5 0.0150
Data collected.
Slope was taken for 1st 100s
y = 0.0022x1.1604
R² = 0.9935
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016
0 1 2 3 4 5 6
Rate
conc/M
Conc/M vs Rate/kPas-1
Without outlier
Power fit taken, y= 0.0022x1.1
To find the order – Rate = k[NaOH]1.1
Order is 1.1

4. Find order NaOH
Let Rate = k[NaOH]x
Rate = k[NaOH]1
1st order respect to NaOH
Expt Conc
NaOH
Initial
rate
1 2 5.04 x 10-3
2 4 10.6 x 10-3
3 5 15.0 x 10-3
Effect of conc NaOH on the rate of hydrogen production
Al + 6H2O + 2NaOH → 2Na[AI(OH)4] + 3H2
𝑅𝑎𝑡𝑒1
𝑅𝑎𝑡𝑒2
=(
𝐶𝑜𝑛𝑐1
𝐶𝑜𝑛𝑐2
)
x
5.04
10.6
=(
2
4
)
x
0.475 = 0.5
x
X = 1.1
Conclusion:
- Min conc NaOH for rxn to happen, 2M
- 1st order with respect to NaOH
1. What is the activation energy for rxn bet AI + NaOH
2. Which catalyst is better, NaOH or KOH?
3. Will reaction with KOH produce the same rate as NaOH
4. What is the order for the rxn bet AI + KOH, will it be 1st order
5. Is it possible for AI to react with HCI instead of NaOH
6. AI will only react with HCI at a certain temp, which is above 60C.
7. What is the activation energy for AI with acid, HCI.
8. Will it be lower or higher than AI with NaOH.
9. To find the purity of AI from various sources, like AI foil, Al metal and AI powder.
Possible research questions
% error =
(𝐿𝑖𝑡 𝑣𝑎𝑙𝑢𝑒−𝐸𝑥𝑝𝑡 𝑣𝑎𝑙𝑢𝑒)
𝐿𝑖𝑡 𝑣𝑎𝑙𝑢𝑒
x 100%
% error =
(1−1.1)
1
x 100%= 10%

5. Graphical Representation of Order :ZERO, FIRST and SECOND order
ZERO ORDER FIRST ORDER SECOND ORDER
Rate – 2nd order respect to [A]
Conc x2 – Rate x 4
Unit for k
Rate = k[A]2
Rate = kA2
k = M-1s-1
Rate
Conc reactant
Rate
Conc reactant Conc reactant
Conc Conc Conc
Time Time Time
Time
Conc reactant
Rate
Time
ln At
Time
1/At
kt
A
A o
t 
 ]
[
]
[
Rate = k[A]0
Rate independent of [A]
Unit for k
Rate = k[A]0
Rate = k
k = Ms-1
Rate vs Conc – Constant
Conc vs Time – Linear
Rate = k[A]1
Rate - 1st order respect to [A]
Unit for k
Rate = k[A]1
Rate = kA
k = s-1
Rate vs Conc - proportional
Conc vs Time
kt
A
A
e
A
A
o
t
kt
o
t


 
]
ln[
]
ln[
]
[
]
[
[A]t
[A]o
kt
A
A o
t


]
[
1
]
[
1
ln Ao
1/Ao
Conc at time t Conc at time t

6. Order of rxn found using THREE mtds
Initial Rate mtd
(Multiple Single Runs)
Conc Vs Time Mtd
(Half Life)
Conc Vs Time Mtd
(Whole Curve/Tangent)
Multiple Single Runs
Vary/Keep certain conc fixed
Wasteful as multiple runs needed
Monitor decrease in conc reactant
Using Half Life to determine order
Monitor decrease conc of single reactant
Using gradient/ tangent at diff conc
Conc x2 – rate x2 - 1st order
Conc x2 – rate x4 – 2nd order
Conc x2 – rate 0 – zero order Convert Conc Vs Time to Rate vs Conc
Rate Vs Conc – Linear – 1st Order
Initial Rate taken, time 0
Draw tangent at time 0
Half Life directly prop to Conc
Half Life inversely prop to Conc
Expt Conc
A
Conc
B
Initial
rate
1 0.01 0.02 2
2 0.01 0.04 4
3 0.02 0.02 4
Conc
Time
Expt 2
Expt 1
Conc reactant
Time
Zero order
Conc reactant
Time
Half Life constant
1st order
2nd order
Conc reactant
Time
Gradient at diff conc
Conc
Rate