2Al(s)+ 3Cl2(g)=2AlCl3(s)
For the above reaction calculate:
A) delta H, then state whether the reaction is endothermic or exothermic
B) delta s, then whether the reaction becomes more or less disorder.
C) delta G, the state whether the reaction is spontaneous or non-spontaneous
To calculate the values requested, we need to use thermodynamic data for the reaction. The standard enthalpy change, ΔH°, standard entropy change, ΔS°, and standard Gibbs free energy change, ΔG°, can be obtained from reference tables or databases. Here are the steps to calculate each value and determine the nature of the reaction:
A) Calculating ΔH° and determining if the reaction is endothermic or exothermic:
1. Find the enthalpy of formation (∆Hf°) values for each compound involved in the reaction. The enthalpy of formation is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states.
- ΔHf° for Al(s) = 0 kJ/mol (by definition, standard enthalpy of elements in their standard states is zero)
- ΔHf° for Cl2(g) = 0 kJ/mol (by definition, standard enthalpy of elements in their standard states is zero)
- ΔHf° for AlCl3(s) = -704.7 kJ/mol
2. Apply Hess's Law, which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual reactions.
ΔH° = Σ (ΔHf° products) - Σ (ΔHf° reactants)
ΔH° = [2ΔHf°(AlCl3)] - [2ΔHf°(Al) + 3ΔHf°(Cl2)]
ΔH° = [2(-704.7)] - [2(0) + 3(0)]
ΔH° = -704.7 kJ
3. Since ΔH° is negative (-704.7 kJ), the reaction releases energy during the formation of the products. Therefore, the reaction is exothermic.
B) Calculating ΔS° and determining if the reaction is more or less disorder:
1. Find the standard molar entropy (∆S°) values for each compound involved in the reaction. Entropy is a measure of the disorder or randomness of a system.
- ΔS° for Al(s) = 28.30 J/(mol·K)
- ΔS° for Cl2(g) = 223.1 J/(mol·K)
- ΔS° for AlCl3(s) = 67.2 J/(mol·K)
2. Apply the same formula as before using the entropy values instead of enthalpy values:
ΔS° = Σ (ΔS° products) - Σ (ΔS° reactants)
ΔS° = [2ΔS°(AlCl3)] - [2ΔS°(Al) + 3ΔS°(Cl2)]
ΔS° = [2(67.2)] - [2(28.30) + 3(223.1)]
ΔS° = 67.2 - 56.60 - 669.3
ΔS° = -658.70 J/(mol·K)
3. Since ΔS° is negative (-658.70 J/(mol·K)), the reaction leads to a decrease in disorder. Therefore, the reaction becomes less disorder.
C) Calculating ΔG° and determining if the reaction is spontaneous or non-spontaneous:
1. Use the equation ΔG° = ΔH° - TΔS°, where T is the temperature in Kelvin (K) and ΔG° is the standard Gibbs free energy change.
2. Assume a temperature value, let's say 298 K (room temperature).
ΔG° = -704.7 kJ - (298 K)(-0.6587 kJ/(mol·K))
ΔG° = -704.7 kJ + 196.8 kJ
ΔG° = -507.9 kJ
3. Since ΔG° is negative (-507.9 kJ), the reaction is spontaneous under standard conditions (at 298 K and 1 atm pressure). Therefore, the reaction is spontaneous.
To answer these questions, we need to first identify the values of ΔH (enthalpy change), ΔS (entropy change), and ΔG (Gibbs free energy change) for the given reaction. Let's go step-by-step:
A) ΔH (enthalpy change):
To calculate the enthalpy change, we need the standard enthalpies of formation for each of the substances involved in the reaction. The standard enthalpy of formation values are:
ΔHf°(Al) = 0 kJ/mol
ΔHf°(Cl2) = 0 kJ/mol
ΔHf°(AlCl3) = -704.8 kJ/mol
Using these values, we can calculate the enthalpy change (ΔH) for the reaction:
ΔH = ΣΔHf°(products) - ΣΔHf°(reactants)
= [2(-704.8 kJ/mol)] - [2(0 kJ/mol) + 3(0 kJ/mol)]
= -1409.6 kJ/mol - 0 kJ/mol
= -1409.6 kJ/mol
Since ΔH is negative (-1409.6 kJ/mol), the reaction is exothermic.
B) ΔS (entropy change):
To determine the entropy change, we need to consider the number of moles of gas (gaseous reactants and products) involved in the reaction.
Reactants:
2Al(s) = 0 moles of gas
3Cl2(g) = 3 moles of gas
Products:
2AlCl3(s) = 0 moles of gas
The net change in the number of moles of gas is (0 moles - 3 moles) = -3 moles.
Entropy change (ΔS) can be calculated using the equation:
ΔS = ΣnΔS°(products) - ΣnΔS°(reactants)
= [0 mol(-26.80 J/(mol·K))] - [3 mol(0 J/(mol·K))]
= 0 J/(mol·K) - 0 J/(mol·K)
= 0 J/(mol·K)
Since ΔS is zero, the reaction does not result in any significant change in disorder.
C) ΔG (Gibbs free energy change):
The Gibbs free energy change is related to ΔH and ΔS through the equation:
ΔG = ΔH - TΔS
Since ΔH = -1409.6 kJ/mol and ΔS = 0 J/(mol·K), we need the temperature (T) to calculate ΔG. Without the value of temperature, we cannot determine whether the reaction is spontaneous or non-spontaneous.