Chemistry
Exercise 6
6. Understand the fundamentals of reaction kinetics and the concerts of energy, enthalpy, entropy and free energy (thermodynamics).
Reaction kinetics is that area of chemistry which concerns itself with the rates of chemical reactions, which factors affect the rate, how the rate is measured and how this can lead to a plausible mechanism of the reaction. Thermodynamics concerns itself with the transformation of energy, the relationship between heat and work, spontaneous and non-spontaneous processes.
a) Ammonia is produced from nitrogen and hydrogen by the Haber-Bosch process:
N2(g) + 3H2 (g) 2NH3(g) ΔH = -92.2 kJ. Would you predict the rate of ammonia formation to increase or decrease:
i) when the partial pressures of N2 and H2 are increased?
Increase
Increasing the partial pressure of the reactants acts to increase the rate of reaction by decreasing the distance between reactants, and thereby increasing the probability that nitrogen and hydrogen will come into contact and react to form ammonia.
ii) when the reaction mixture is cooled?
Decrease
Cooling the reaction mixture decreases the kinetic energy of the reactants, which reduces the probability of the reactants coming together with sufficient energy to overcome the activation barrier. The result is a decrease in the rate of formation of ammonia
iii) when a catalyst is added?
Increase
A catalyst acts on a system by reducing the activation energy required for reactants to form products. In the absence of a catalyst, not all collisions between nitrogen and hydrogen molecules occur with sufficient energy or in the proper orientation. By reducing the required activation energy, catalysts increase the probability that the reactants will come together to form a product.
iv) Would the value of the equilibrium constant increase, decrease or remain the same in each of these situations?
i) Stay the same – only temperature changes affect the equilibrium constant.
ii) Increase – since the reaction is exothermic a decrease in temperature will favour the forward reaction. This increases the amount of ammonia leading to an increase in K.
iii) the presence of a catalyst acts to increase the rate of reaction, but it does not change the position of the equilibrium; therefore the equilibrium constant remains the same.
b) The enthalpy (heat) of combustion of ethanol is 1368 kJ/mol.
i. Write a balanced equation for the combustion of ethanol.
All combustion reactions of organic molecules involve oxygen as a reactant and carbon dioxide and water as products. The balanced reaction is
2 CH3CH2OH + 6 O2 
4 CO2 + 6 H2O
ii. When 23.0 g of ethanol is combusted and all the heat released is transferred to 10.0 l (10.0 kg) of water, initially at 20.0 C, to what temperature does the water heat up? The specific heat for water is 4.18 J/g. C.
This problem can be solved by determining the amount of energy in joules produced in the combustion of 23g of ethanol and the resulting increase in water temperature in the presence of released energy.
1. We are given the heat of combustion for ethanol per mole (-1368 kJ/mol – the negative indicates that heat is given off), so the first thing to do is to calculate the number of moles burned.
Molar massCH3CH2OH = 46 g/mol
molesCH3CH2OH = 23 g/ 46 g/mol = 0.5 mol
2. Determine the amount of energy given off during combustion using the following formula:
Q = n* Hc
Q = 0.5 mol * (-1368 kJ/mol) = -684 kJ/mol
3. Assuming that all the energy from the combustion of ethanol is transferred to the water, the increase in temperature can be calculated using the formula
Q = mcΔT,
where Q is the amount of energy in joules, m is the mass of water in grams, c is the specific heat capacity of water and ΔT is the change in temperature.
Solving for ΔT,
4. Determine the final temperature, Tf = Ti + ΔT
Tf = 20oC + 16.4oC = 36.4oC
iii. What would the final temperature of the water be if 23.0 g of propane (ΔHcomb = -2220 kJ/mol) was combusted?
This problem is exactly the same as the above, except in this case it is propane (CH3CH2CH3) that is burned, rather than ethane.
Repeating the calculation using the molar mass of propane which is 44.1 g/mol. The final temperature is 47.8 C.
c) Calculate the enthalpy change ΔH and the entropy change ΔS of the reaction C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l) from the enthalpies of formation and absolute entropies of reactants and products. What is the value of the free energy change ΔG at 25oC? Is the reaction spontaneous or non-spontaneous? Is there a relationship between spontaneity and rate?
To solve this, you will need to reference the standard enthalpies of formation and entropies found in any introductory chemistry textbook, keeping in mind that all elements in their standard states are given an Ho of zero.
| ΔH | ΔS | |
| C6H12O6(s) | -1268 kJ/mol | 212 J/K*mol |
| O2(g) | 0 | 205 J/K*mol |
| CO2(g) | -393.5 kJ/mol | 214 J/K*mol |
| 6H2O(l) | -285.8 kJ/mol | 70 J/Kmol |
The enthalpy and entropy changes for the above reaction can be calculated using the following formula:
and
Now that ΔH and ΔS for the reaction have been calculated, The Gibb’s Free Energy (ΔG) can be determined using the following formula; ΔG= ΔH - TΔS, where temperature is given in Kelvins
Solving for ΔG gives,
ΔG= (-2809 kJ/mol) – (298 K * 262 J/K*mol)
ΔG= (-2809 kJ/mol) - (+78076 J/mol)
= (-2809 kJ/mol) - (+78.1 kJ/mol)
= - 2731 kJ/mol
The sign of the Gibb’s free energy value is negative, thereby indicating that the reaction is spontaneous (a non-spontaneous reaction has a ΔG value greater than zero, and a ΔG value of 0 indicates that the reaction is at equilibrium). While thermodynamics can predict whether a reaction is spontaneous, (i.e. energetically favourable) it cannot predict the rate at which the reaction will go forward.
