Study the reaction kinetics and find out evidence about the mechanism between the reaction of acid and magnesium.
Investigation of acid and magnesium
This investigation is to study the reaction kinetics and find out evidence about the mechanism between the reaction of acid and magnesium. It can be done by finding out the rate, order, activation energy and enthalpy change of the reaction.
Acid
What is an acid? Definition: an Arrhenius acid is a compound containing hydrogen which will form hydrogen ions in water.
What is a strong acid? Acids that donate their acidic protons almost completely to water are classified as strong acids; acids that interact very little with water are classified weak acids.
Ref, 'Nuffield Chemistry Students' Book'
Ionic equation for acid and Mg:
2H+(aq) + Mg(s) --> H2(g) + Mg2+(aq)
Rate equation and order of reaction
Reactants are converted to products during a reaction and the rate of reaction is a measure of how fast the reaction is. It is an indication of how much reactants are converted into product in a given time. It can be affected by the concentration. The rate equation can be expressed as:
Rate ? [A] a [B] b
Rate = k [A] a [B] b
Rate = k[Acid] a
N.B.[Mg] is not needed in the rate equation: because Mg is a solid, so its concentration does not vary.
k is the rate constant
a is the order of reaction
Order with respect to any one reactant is the index. The sum of the order of reaction is calculated by add the index of the reactants, in this case, is only the concentration of acid [acid].
Order means how much dependence of the rate is on the concentration of the reactant.
i.e. n is usually 0,1,2 when n=0, so [a]0=1, so Rate = k ; when n=1,Rate = k[a] ; when n =2, Rate = k [a]2
From the rate equation, we can see that [acid] involve in the Rate Determining Step (that means it is the slowest step within the reaction), because the rate depends on [acid]. We find the rate equation through experiment, then find out the order.. Together we can deduce its mechanism.
Activation energy, EA
Activation energy is the energy needed for the reactants to reach some inter-mediate state. This tells us how likely the reaction would happen.
Particles have to have sufficient energy(greater than EA). In the case, the energy is mostly kinetic energy, since ke=1/2 mv2, and the distribution of speeds can be determined by Zartmann, so the energy distribution is determined as the graph below.
Enthalpy change, ?H
Definition: standard enthalpy change of a reaction as the amount of heat absorbed or evolved when the molar quantities of reactants as stated in the equation react together under standard conditions, i.e. at pressure of 1 atm, temperature of 298K, with substances in their normal physical states under these condition and solution having unit activity.
Enthalpy is the changes in the heat content. These changes can be shown through the change in temperature.
?H= Enthalpy after - enthalpy after
Enthalpy change of a reaction sometimes can be a rough guide to the likelihood that the reaction will occur. We can see if ?H is negative, hence energy lost, so the product is more stable, means the reaction would happen.
Ref, 'Chemistry in context'
Integrated Rate Laws
The rate of reaction is proportional to the rates of change in concentrations of the reactants and products; that is, the rate is proportional to a derivative of a concentration.
Rate equation is a differential equation that relates the rate of change in a concentration to the concentration itself. Integration of this equation produces the corresponding integrated rate law, which relates the concentration to time.
Reaction Order
Differential Rate Law
Integrated Rate Law
Characteristic
Kinetic Plot
Slope of
Kinetic Plot
Units of
Rate Constant
Zero
-
d [A]
d t
= k
[A] = [A]0 - k t
[A] vs t
- k
mole dm-1 sec-1
First
-
d [A]
d t
= k [A]
[A] = [A]0 e- k t
ln [A] vs t
- k
sec-1
Second
-
d [A]
d t
= k [A]2
[A] =
[A]
+ k t [A]0
/[A] vs t
k
Dm mole-1 sec-1
Ref:'http://www.chm.davidson.edu/ChemistryApplets/kinetics/IntegratedRateLaws.html'
The second order can be also rearranged to 1/[A] - 1/[A]0 = kt
Ref: http://www.cartage.org.lb/en/themes/Sciences/Chemistry/Miscellenous/Helpfile/Kinetics/secondorder.htm
Method
Two acids I have chosen are Hydrochloric acid HCl and Sulphuric acid H2SO4, one monobasic acid and one dibasic acid.
Planning
Order and Rate(experiment 1)
The product are hydrogen gas and a salt, and gas can be easily measured and collected by using a syringe, and the rate of the reaction is proportional to the gas produced, so by know how much gas is produced per unit time, rate can be found.
It can be done by two methods: Continuous Rate Method, or Initial Rate method. Continuous Rate Method involves tracking how much gas is produced at a certain time interval until the reaction has stopped. Initial Rate Method is only to measure the 10% to 20% of the reaction where we can assume that the concentration of the reactant has not changed.
I was provided with acids with 2M concentration, so I had to dilute myself. Mg comes with ribbon or powder.
Preliminary work
I tried with Continuous Rate Method, because it seems would shows me how the rate very with time.
Mg can either be powder or ribbon, I choose ribbon because I can see the reaction more clearly and the rate would not be too fast.
The reaction for HCl is:
2HCl(aq) + Mg(s) H2(g) + MgCl2 (aq)
2x0.02 0.02 0.02
0.02x24=0.48g
20 cm3 480ml
I used 20cm3 of acid solution. It was my initial guess of the acid quantities needed. I used 100% excess of Mg, 0.96g. It produced 100ml ...
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I tried with Continuous Rate Method, because it seems would shows me how the rate very with time.
Mg can either be powder or ribbon, I choose ribbon because I can see the reaction more clearly and the rate would not be too fast.
The reaction for HCl is:
2HCl(aq) + Mg(s) H2(g) + MgCl2 (aq)
2x0.02 0.02 0.02
0.02x24=0.48g
20 cm3 480ml
I used 20cm3 of acid solution. It was my initial guess of the acid quantities needed. I used 100% excess of Mg, 0.96g. It produced 100ml of gas in 3 seconds, (in theory it would produced, 480ml of gas), so 21% of the reaction. That is too fast. I did another experiment with 1M acid solution with 100% excess Mg,0.48g, 100ml of gas only takes 5 seconds. That is too fast too.
I adjusted the solution to 10 cm3, it is the minimum value, because less than 10 cm3 the Mg can not be excess. For 2 moldm-3 acid with 10 cm3 and excess Mg would produced 240ml of gas and the syringe is only 100ml, which is only 42% of the reaction, so Continuous Rate Method is not good, also, I found that the method was very difficult to do and to get an accurate set of result, because having to keep timing and recording result include a high percentage of experimental error.
Acid volume measurement is important, and measuring cylinder was not accurate enough, so I decided to use burette.
Mg has to be an excess to not affect the rate, and the more percentage of excess, the less error it would be. Surface area has to be controlled.
Experiments 1--3
Safety:
2 moldm-3 acids are corrosive, so wear glove.
Hydrogen is explosive, so keep the room well ventilated (i.e. keep windows open) and keep the gas well away from Bunsen Burner.
Equipments:
Burette x2
Side arm test tube x1
Glass syringe x1
Stopwatch x1
Thermometer x1
Conical Flask x1
Experiment 1
I decided to do Initial Rate Method, because it is during in the range of the syringe and the rate can be adjusted to be slower, so easier to be measured. I time how long 20% of the reaction would be and the rate would be 1/time.
0cm3 of the acid solution is used every time. I am only collecting 20% of the gas, so only 20% of the Mg is only used in the reaction, 80% is left as excess, it has 400% of excess, which minimize the error and I use the same types of ribbon of Mg and coil them into circle.
Diagram:
Calculation for [acid] and Mg used
Table 1
HCl and Mg
Concentration/moldm-3
Acid/cm3
Distilled water/cm3
Mole
Total Gas/ml
20% gas/ml
Mg/g
Time/s
0.25
.25
8.75
0.00
60.00
2.00
0.03
0.50
2.50
7.50
0.01
20.00
24.00
0.06
0.75
3.75
6.25
0.01
80.00
36.00
0.09
.00
5.00
5.00
0.01
240.00
48.00
0.12
.25
6.25
3.75
0.01
300.00
60.00
0.15
.50
7.50
2.50
0.02
360.00
72.00
0.18
.75
8.75
.25
0.02
420.00
84.00
0.21
2.00
0.00
0.00
0.02
480.00
96.00
0.24
Table2
Concentration/moldm-3
Acid/cm3
Distilled water/cm3
Mole
Total Gas/ml
20% gas/ml
Mg/g
Time/s
0.25
.25
8.75
0.00
60.00
2.00
0.06
0.50
2.50
7.50
0.01
20.00
24.00
0.12
0.75
3.75
6.25
0.01
80.00
36.00
0.18
.00
5.00
5.00
0.01
240.00
48.00
0.24
.25
6.25
3.75
0.01
300.00
60.00
0.30
.50
7.50
2.50
0.02
360.00
72.00
0.36
.75
8.75
.25
0.02
420.00
84.00
0.42
2.00
0.00
0.00
0.02
480.00
96.00
0.48
H2SO4 and Mg
N.B. Reaction for H2SO4 used twice amount of Mg, because HCl : Mg is 2:1 and H2SO4 : Mg is 1:1
Result All values are in 2 decimal places
Table 3 HCl and Mg
Concentration/moldm-3
Time/s
Repeat Time/s
Time/s
Rate(1/s)
0.25
292.00
292.00
0.00
0.50
87.66
87.66
0.01
0.75
38.58
38.58
0.03
.00
5.37
5.37
0.07
.25
1.85
1.85
0.08
.50
2.54
9.47
9.47
0.11
.75
7.34
7.34
0.14
2.00
0.21
7.30
7.30
0.14
Table 4
Concentration/moldm-3
Time/s
Repeat time/s
Time/s
Rate(1/s)
0.25
78.43
58.51
58.51
0.02
0.50
4.34
4.34
0.07
0.75
4.70
0.54
0.54
0.09
.00
8.22
8.22
0.12
.25
6.84
6.84
0.15
.50
4.52
4.52
0.22
.75
5.60
3.53
3.53
0.28
2.00
3.40
3.40
0.29
Some the results do not fit into the pattern (time get shorter as [acid] increase), so I did a repeat on them. The reason for the wrong results may because I did parts of my experiment in other lesson, which means the syringe was stiff to start with and the temperature in the room differ.
Activation energy(experiment 2)
According to the Arrhenius equation:
Rate constant (k) = A exp(-EA/RT)
A a constant, number of collision per second
EA Activation energy (J mole-1)
R Gas constant (8.31 J mol-1 K-1)
T Temperature (K)
Ref, 'Chemistry in context'
Nature log the equation we get:
Ln k = - EA/RT + Ln A
Therefore, if measure rate constant at various temperture, so Activation energy can be found.
N.B rate constant is hard to find, because Rate = k[Acid] a
Rate ? k , so we can Ln rate instead of Ln Rate Constant
I varied the temperature from 20 oc to 70 oc. I kept the mass of Mg and concentration of the acid constant.
The reaction is very exothermic, so I took the temperature before, after and get the average, to reduce error.
Concluded my preliminary work, I decided to use 0.06g of Mg and 0.75M, 20cm3 HCl, because 0.06g is 0.0025mole (0.06/24), then 0.0025 x24000= 60ml, produce 60ml of gas, that is within the limit of the syringe, and 0.75M and 20cm3 HCl would give me a steady rate, so easy to measure.
For H2SO4, I used 0.06 g of Mg and 1M, 10cm3 H2SO4, the reason is same as above.
Experiment 2
Diagram:
Result:
Acid
Acid/ cm3
Distilled Water/ cm3
HCl
7.5
2.5
H2SO4
7.5
2.5
Table 5 HCl and Mg
Temp before/oc
Temp after/oc
Temp average/oc
Time/s
Rate(1/s)
Temp/K
/T(1/K)
Ln Rate
23.00
31.00
27.00
72.62
0.01
300.00
0.00
(4.29)
31.00
38.00
34.50
65.59
0.02
307.50
0.00
(4.18)
40.00
43.00
41.50
54.60
0.02
314.50
0.00
(4.00)
51.00
52.50
51.75
43.04
0.02
324.75
0.00
(3.76)
60.00
64.00
62.00
38.06
0.03
335.00
0.00
(3.64)
69.00
73.00
71.00
31.30
0.03
344.00
0.00
(3.44)
Table 6 H2SO4 and Mg
Temp before/oc
Temp after/oc
Temp average/oc
Time/s
Rate(1/s)
Temp/K
/T(1/K)
Ln Rate
22.00
37.00
29.50
73.58
0.01
302.50
0.00
(4.30)
30.00
41.00
35.50
60.52
0.02
308.50
0.00
(4.10)
40.00
49.00
44.50
54.31
0.02
317.50
0.00
(3.99)
50.00
53.00
51.50
45.28
0.02
324.50
0.00
(3.81)
61.00
63.00
62.00
38.24
0.03
335.00
0.00
(3.64)
69.00
68.00
68.50
37.60
0.03
341.50
0.00
(3.63)
Enthalpy (experiment 3)
Diagram:
I used 0.06g of Mg react with 10cm3 of acid in a test tube. I measured the temperature before and the maximum temperatures, which is when the reaction finished.
Results:
Table 7 HCl and Mg
Concentration/moldm-3
Temp Before/oc
Temp after/oc
Change in temp/oc
.00
23
50
27
.50
24
51
27
2.00
22
50
28
Table 8 H2SO4 and Mg
Concentration/moldm-3
Temp Before/oc
Temp after/oc
Change in temp/oc
.00
23
50
27
.50
22
54
30
2.00
24
55
31
Then I realized that the enthalpy change should be the same for each concentration, because the same amount of Mg has reacted. But the low concentration has a lower rate, so it would lose more heat to the surrounding during the reaction that may explain why the change in temperature is slightly lower.
So I suggest the change in temp for the HCl is 28 oc and for H2SO4 is 31 oc, simply just take the highest value.
Analysis:
Uncertainty
Experiment 1 has measuring error, i.e. reading on the syringe, they are +or-1ml out of 100ml, that is 2%, the Mg is 400% excess, is about 5% error??(roughly)explain, the reading on the thermometer is +or-1,that is 2%.
So for experiment 1, error is roughly 2%(syringe) +5%(Mg excess) is 7%
For experiment 2, error is roughly 2%(syringe)+2%(temp before)+2%(temp after)=6%
For experiment 1:
The reaction for HCl is:
2HCl(aq) + Mg(s) H2(g) + MgCl2 (aq)
Table9 Effect of [HCl] on rate (Mg in excess)
Concentration/moldm-3
Time/s
Rate(1/s)
Log(HCL)
Log rate
Con2 (mol2dm-6)
0.25
292.00
0.00
-0.60
-2.47
0.06
0.50
87.66
0.01
-0.30
-1.94
0.25
0.75
38.58
0.03
-0.12
-1.59
0.56
.00
5.37
0.07
0.00
-1.19
.00
.25
1.85
0.08
0.10
-1.07
.56
.50
9.47
0.11
0.18
-0.98
2.25
.75
7.34
0.14
0.24
-0.87
3.06
2.00
6.78
0.15
0.30
-0.83
4.00
Table 10 Effect of [H2SO4] on rate (Mg in excess)
Concentration/moldm-3
Time/s
Rate(1/s)
Log(Con)
Log Rate
Con2 (mol2dm-6)
0.25
58.51
0.02
-0.60
-1.77
0.06
0.50
4.34
0.07
-0.30
-1.16
0.25
0.75
0.54
0.09
-0.12
-1.02
0.56
.00
8.22
0.12
0.00
-0.91
.00
.25
6.84
0.15
0.10
-0.84
.56
.50
4.52
0.22
0.18
-0.66
2.25
.75
3.53
0.28
0.24
-0.55
3.06
2.00
3.00
0.33
0.30
-0.48
4.00
Graph 1 using table 10, I plot Rate against [HCl] and found the line of best fit is not a straight line, and looks like an curve, this means the order is not first order.
Graph2 using table 11, I plot Rage against [HCl]^2 and found the line of best fit is almost a straight line, and go through the origin, but some points are off the line, they may due to experimental errors. It is suggest the order of this reaction is roughly 2.
(because I squared the [HCl], so it means the uncertainty is twice, so 14% of uncertainty)
From graph 10 and 11, the last two results are not very well fitted with the line of best fit, that suggest the rate decrease at high concentration, but this can not be the case. However, consider from the other side, there was a time lap between when Mg is in the solution and placed the bung, [acid] would decrease rapidly (from our prediction that high [acid], high rate), that suggest we may measured a lower [acid]'s rate. However, it can also means at high concentration, the rate slow down.
Uncertainty
The error bars in graph 1 and 2 suggest the uncertainty is higher than I purposed, because when I tried to draw a line of best fit, the error bars have not cover the line.
I used another method to find the exact order.
Because Rate = k[Acid] a
Therefore Log Rate = a log[HCl] + log k
Y = m X + c
The equation is
y = 1.90x - 1.31
Order respect to [HCl] is 1.90 (3sf), that is also the overall order. This means the concentration of [HCl] against time graph is a very deep curve and the half-life is not constant but will increase rapidly as the reaction proceeds.
Log k = -1.31, so k = 10-1.31 = 0.0490 (3sf)
N.B unit for rate constant is not definite, because I am not certain what order the reaction is
So Rate = 0.0490 [HCl] 1.90
This shows the rate is very dependent on the [HCl], the concentration against time is illustrated in graph 8.
The reaction for H2SO4 is:
H2SO4(aq) + Mg(s) H2(g) + Mg SO4 (aq)
Graph 4 is Rate against [H2SO4]. I found the line of best fit is roughly a straight line. Graph 5, I plot Rage against [H2SO4] 2 and found the line of best fit is a curve.
N.B the error bars shows the uncertainty is higher than I purposed
Using the log method again, I found the line is y = 1.33x - 0.889
So overall order is 1.33 and k is 0.129
So Rate = 0.129 [H2SO4] 1.33
However, the order of a reaction must be a integer, because number of particles are discrete. So the order I obtain must be an average order of the reaction.
From graph 3 and 6, the points are not fitted with the line of best fit that means the gradient is not constant, hence the order is not constant throughout the reaction. The gradient of the first few points is big and decrease hence the order is high at the beginning, and decrease as the reaction proceeds.
Integrated Rate Laws
Using the integrated Rate Laws, I can see how [acid] varies with time. I assume [HCl] is second order and [H2SO4] is first order.
For [HCl] Rate = 0.0490 [HCl] 1.90
According to the equation: 1/[A] - 1/[A]0 = kt
/ [HCl] - 1 / [2M] = 0.048 x t
/ y - 1 / [2M] = 0.048 x x
y = 1/ (0.048 x +0.5 ) graph 9
The results correspond to table 3 very well.
For [H2SO4] Rate = 0.129 [H2SO4] 1.33
According to equation: [A] = [A]0 e- k t
[H2SO4] = [2M] exp(-0.129t)
y = 2 exp(-0.129x) graph 10
According to the graph, the reactions seems stop after 50seconds, which is the same results in table 4.
However, these two graphs are only estimations, because I have to assume the order of the reaction in the first place.
For Experiment 2:
HCL and Mg
Using table 6, I plotted graph 7. From page 8, we know Ln k = - EA/RT + Ln A
y = -2000x + 2.37
so the gradient is -2000, correspond to - EA/R, so -2000xR=-2000x8.31=-16620j=- EA
So EA = 16620J=+16.6kJ(3sf)
H2SO4 and Mg
Using table 7, I plotted graph 8, do some calculation as above, line of best fit is
y = -1800x + 1.71
EA= 1800x8.31=15000j=+15kJ
According to the rate equation, Rate = k[Acid] a, a would affect the rate, and I used Ln K for the calculation above, where I assume a is a constant. My results from experiment 1 shows that the order may have changed as concentration changes , according to graph 1 and 2, because the rate slowing down. Just look at the last two results on graph 1 and 2, it seems the order has decrease to zero, because the line of best fit is a horizontal line. This can due to the experiment error I described in page 12.
Same case applied sulphuric acid, the order may has slow down as concentration increases.
For experiment 3
Enthalpy
HCl and Mg
Q=cmT
= specific heat capacity of solution x mass of the solution x temperature change
?H
= - 4.18 x 10 x 28 = - 1170.4 j = -1.17kJ
H2SO4 and Mg
Q=?H
= - 4.18 x 10 x 31 = - 1.30 kJ
Having work out the enthalpy change for the whole reaction, the energy liberated during 20% of the reaction can be calculated.
Table 11 Change of temperature during the 20% of the reaction for HCl and Mg
Concentration/ moldm-3
Volume/ cm3
Mole
Energy liberated during the 20% reaction/j
Change in Temp/ oc
0.25
0.00
0.00
0.59
0.01
0.50
0.00
0.01
.17
0.03
0.75
0.00
0.01
.76
0.04
.00
0.00
0.01
2.34
0.06
.25
0.00
0.01
2.93
0.07
.50
0.00
0.02
3.51
0.08
.75
0.00
0.02
4.10
0.10
2.00
0.00
0.02
4.68
0.11
Table 12 Change of temperature during the 20% of the reaction for H2SO4 and Mg
Concentration/ moldm-3
Volume/ cm3
Mole
Energy liberated during the 20% reaction/j
Change in Temp/ oc
0.25
0.00
0.00
0.65
0.02
0.50
0.00
0.01
.30
0.03
0.75
0.00
0.01
.95
0.05
.00
0.00
0.01
2.60
0.06
.25
0.00
0.01
3.25
0.08
.50
0.00
0.02
3.90
0.09
.75
0.00
0.02
4.55
0.11
2.00
0.00
0.02
5.20
0.12
The material below is reference to 'Chemistry in Context', it talks about the effect of temperature rise(below)
From the kinetic theory, we can predict the relative increase in number of collisions when the temperature rises by 10K. The kinetic energy of a particle is proportional to its absolute temperature:
/2mv² ? T
But the mass of a given particle remains constant
V² ? T
Therefore V²1 = T1
V²2 T2
Where V1 is the velocity at temperature T1, and V2 is the velocity at temperature T2.
Now, suppose that the average speed of a particle is V at 300K. We can work the average speed at 310K by using the equation,
V²1 = 310
V² 300
Therefore V1= 310
300
V= 1.033V
= 1.016V
This therefore shows that the average speed at 310K is only 1.016 times greater than that at 300K i.e. it has only increased by 1.6%.
So as the calculation above, I can work out the increase in speed of particles, using the temperature change in table 12 and 13.
Temperature in the solution is 24 oc, as I measured.
Table 14 % increase in velocity of the particles during the reaction for HCl and Mg
Concentration/ moldm-3
Temp before(T1)/ oc
Temp after(T2)/ oc
T2/T1
% increase in V
0.25
297.00
297.01
.00
0.00
0.50
297.00
297.03
.00
0.00
0.75
297.00
297.04
.00
0.01
.00
297.00
297.06
.00
0.01
.25
297.00
297.07
.00
0.01
.50
297.00
297.08
.00
0.01
.75
297.00
297.10
.00
0.02
2.00
297.00
297.11
.00
0.02
Table 15 % increase in velocity of the particles during the reaction for H2SO4 and Mg,
Concentration/ moldm-3
Temp before(T1)/ oc
Temp after(T2)/ oc
T2/T1
% increase in V
0.25
297.00
297.02
.00
0.00
0.50
297.00
297.03
.00
0.01
0.75
297.00
297.05
.00
0.01
.00
297.00
297.06
.00
0.01
.25
297.00
297.08
.00
0.01
.50
297.00
297.09
.00
0.02
.75
297.00
297.11
.00
0.02
2.00
297.00
297.12
.00
0.02
Entropy
The entropy of this reaction also tell us how likely the reaction would occur.
HCl and Mg
2HCl(aq) + Mg(s) --> H2(g) + MgCl2 (aq)
?Stotal=?Ssurrouding+ ?Ssystem
?Ssurrouding= -?H/T, (T=298K, room temp) and we worked out ?H before, -1.17kJ, ??Ssurr = +3.93 Jmol-1K-1
?Ssystem=Product - Reactant , values are obtained from the databook,
H2 (65.3x2)+MgCl2 (89.6)- 2(H+(0)+Cl-(56.5))-Mg(32.7)=+74.5 Jmol-1K-1
?Stotal= +78.43 Jmol-1K-1
We can deduced that the reaction would happen spontaneously in 298K
H2SO4 and Mg,
H2SO4 (aq) + Mg(s) --> H2(g) + Mg SO4(aq)
Same calculation as above, ?Stotal=?Ssurrouding+ ?Ssystem
?Ssurr=+3.79 Jmol-1K-1
?Ssys= Mg SO4 (91.6) + H2 (65.3x2) - Mg(32.7) - H2SO4 (0+ 20.1)=+169.2 Jmol-1K-1
?Stotal= +173 Jmol-1K-1
Confidently, we can deduced that the two reaction would happen spontaneously in 298K
So far, for HCl and Mg, I have found
Rate = 0.0490 [HCl] 1.90
EA =+16.6kJ
?H= - 1.17kJ per mole
?Stotal= +78.43 Jmol-1K-1
For H2SO4 and Mg,
Rate = 0.129 [H2SO4] 1.33
EA= +15kJ
?H= - 1.30kJ per mole
?Stotal= +173 Jmol-1K-1
The likelihood of the reaction
The reaction is very likely to happen. This is because ?Stotal is positive, suggest spontaneous reaction and its =+16.6kJ is not very high. When the concentration increases, the temperature increases as in graph 10, hence a increase in speed of particles (graph 12), and 0.5xMxv2 is KE, so KE increase by the square of v, which is a big gain in energy, so the activation energy curve is shifted to the right hand said and more particles has the KE to go over the energy barrier. This means the higher the concentration the higher increase in temperature, so higher velocity of particles, higher KE, hence more particles have enough energy to complete the reaction, so more likely the reaction would happen.
Difference in rate
Two acids are ionic
Ionic equation for acid and Mg:
2H+(aq) + Mg(s) --> H2(g) + Mg2+(aq)
H2SO4 is dibasic and HCl is monobasic. They are all strong acid, means the H+ would dissociated fully. H2SO4 has 2 acid protons and HCl only has 1 acid proton. So 0.1mol wrt [HCl], is 0.1mol wrt [H+], and 0.1mol wrt [H2SO4] and 0.2mol wrt [H+]. The rate for sulphuric acid ought to be higher than hydrochloric acid, because the more the acid proton, the quicker the acid proton dissociate.
However, when we consider about the spectator ions, SO42+ and Cl-, there is a significant difference in size, Cl- has ionic radius of 0.180nm, S8+ has ionic radius of 0.102nm, covalent radius of 0.102nm, and O2+ has ionic radius of 0.140nm and covalent radius 0.073nm.
For SO42+ overall is roughly about 0.140
0.103 0.101
0.140
the length of SO42+ is about 0.140 + 0.102 + 0.102 = 0.344 is about 2 times bigger the Cl ions. The atomic weight of SO4 is 32 + 16x4 = 96, and Cl is 35. SO4 is 2.7 times heavier.
From these evidence, I can suggest that during the collision, the spectator ions SO4 and Cl affect the rate of reaction as well. The SO4 ion is bigger than the Cl, so the sulphuric acid reaction, the SO4 slows down the rate of collision because of its size. But for hydrochloric aicd reaction, the Cl is relatively small and lighter, so the collision is easier to happen, hence a quicker rate of reaction.
Keith Li
11
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