IA on effect of temperature of HCI on the rate of hydrogen production, and finding Ea for reaction between AI with HCI.

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 temp of HCI on the rate of hydrogen production, and finding Ea for reaction
Hydrogen is used for fuel cell.
Aluminium is reactive – but will not react with acid as it forms a stable aluminium oxide layer.
HCI will react only at certain temp, around 60C. Source of H2 using aluminium.
Finding the activation energy.
AI foil will be tested
2AI + 6HCI → 2AI(CI)3 + 3H2
Reaction mechanism
Procedure:
5ml 1M HCI is incubated at various temp for 5 mins.
Temp correction was done.
0.01g AI foil was added to 5ml 1M HCI.
Temp HCI – 65C, 70C, 75C, 80C, 85C.
Pressure sensor will be used to measure the initial rate rxn.
Slope of pressure change over time – used as initial rate.
Assume conc HCI in excess, doesnt change much over time.
Al powder/metal

2. HCI at 65C Rxn happen at 1M HCI
Pressure sensor to measure rate
Temp
C
Rate
kPa/s-1
65 0.005617
70 0.01517
75 0.08274
80 0.1641
85 0.2634
Data collected.
Slope was taken for 50-100s Temp increases, rate increases exponentially
Effect of temp of HCI on the rate of hydrogen production, and finding Ea for reaction
y = 1E-08e0.2015x
R² = 0.9553
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
50 60 70 80 90
Rate temp/C
Temp/C vs Rate
Negative control – without acid, AI will not produce any H2 gas at 65C.
Slight increase in pressure due to increase in temp, as PV=nRT. Increase in pressure is insignificant.

3. Arrhenius Eqn
Ea from its gradient
Arrhenius Eqn - Ea by graphical Method
RT
Ea
e
A
k


 .
. 







T
R
E
A
k a 1
ln
ln
Plot ln k vs 1/T
ln both sides
-Ea/R
Gradient = -Ea/R
Gradient = - 25188
-25188 = -Ea/R
Ea = 25188 x 8.314
= 209kJmol-1
ln k
1/T
Temp/K 1/T k ln k
338 2.95 x 10-3 0.005617 -5.181
343 2.91 x 10-3 0.01517 -4.188
348 2.87 x 10-3 0.08274 -2.492
353 2.83 x 10-3 0.1641 -1.807
358 2.79 x 10-3 0.2634 -1.334
Arrhenius plot to find Ea.
Rate = k[HCI]1
. 1st order. Conc HCI is 1M, assuming doesn’t change much/excess
rate constant, k. = Rate. Rate = k [1M]1
Ea = 209 k Jmol-1
Lit value Ea.= (55-60 kJmol-1). Click here, here, here.
% error =
(𝐿𝑖𝑡 𝑣𝑎𝑙𝑢𝑒−𝐸𝑥𝑝𝑡 𝑣𝑎𝑙𝑢𝑒)
(𝐿𝑖𝑡 𝑣𝑎𝑙𝑢𝑒)
x100%
% error =
(60 −209)
(60)
x100%=248%
2AI + 6HCI → 2AI(CI)3 + 3H2
y = -25188x + 69.288
R² = 0.9553
-6
-5
-4
-3
-2
-1
0
0.00275 0.0028 0.00285 0.0029 0.00295 0.003
lnk
1/T
lnk vs 1/T

4. Temp and rate constant link by Arrhenius Eqn
X + Y → Z
Rate of rxn = (Total number collision) x ( fraction collision, energy >Ea) x ( [X] [Y] )
Arrhenius Constant
A
Fraction molecule energy > Ea
e –Ea/RT
Conc
[X][Y]
Rate of rxn = A e –Ea/RT [X][Y]
Rate of rxn = k [X] [Y]
If conc constant BUT Temp changes, combine eqn 1 and 2
Rate of rxn = k [X]1
[Y]1
= A e –Ea/RT [X][Y]
k = A e –Ea/RT
Rate rxn written in TWO forms
Rate of rxn = A e –Ea/RT [X] [Y]
Eqn 1 Eqn 2
Cancel both sides
Arrhenius Eqn - Ea by graphical Method
RT
Ea
e
A
k


 .
.








T
R
E
A
k a 1
ln
ln
Plot ln k vs 1/T
• Gradient = -Ea/R
• ln A = intercept y axis
ln both sides
ln k
1/T
-Ea/R
1. Determine Ea for 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 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. Determine Ea for AI + 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.
10. To investigate if AI foil will react with food if placed in oven, or use in BBQ.
If AI reacts with food, the amt of AI will be lower, and it can be quantified using NaOH or KOH
11. Determine Ea for AI foil, AI metal, AI powder, and AI of various thickness.
Research Questions