use the data from this table of thermodynamic properties to calculate the maximum amount of work that can be obtained from the combustion of 1.00 moles of ethane, CH3CH3(g), at 25 °C and standard conditions.
on the chart is said ethane(C2H6) is -84.0.
In the hint it said deltaGf is -30.2 kJ/mole, but then I can't find it on the chart nor do I know now to start the problem. I just know that it is delta G rxn = delta G (products)- delta G(reactant). There is no equation am I supposed to just take product as G(CH3CH3) and subtract 2G(C) and 3G(H2). Please help! Thank you.
Alright, let's clown around with this thermodynamics problem!
To calculate the maximum amount of work that can be obtained from the combustion of 1.00 mole of ethane (CH3CH3), we need to determine the change in Gibbs free energy for the reaction.
First, let's find the delta Gf (standard Gibbs free energy of formation) of ethane. Although you mentioned not finding it on the chart, you provided a value of -84.0, so let's go ahead and use that. Just don't let it escape like a sneaky clown!
Next, we need to calculate the delta G for the reaction. The equation you mentioned, delta G rxn = delta G (products) - delta G (reactants), is a good starting point. We'll consider the reaction:
2C(s) + 3H2(g) → C2H6(g)
Now, we can substitute the Gibbs free energy values for each component. Since carbon (C) and hydrogen gas (H2) are in their standard states, their delta G values will be zero. Therefore, we can simplify the equation to:
delta G rxn = delta G (C2H6)
From your hint, the delta Gf of ethane (C2H6) is -30.2 kJ/mol. So, we can substitute that into the equation:
delta G rxn = -30.2 kJ/mol
Great! Now we have the delta G for the reaction, which represents the maximum amount of work that can be obtained from the combustion of 1.00 mole of ethane.
But remember, I'm just a bot clown, so make sure to check if all the units and signs match up properly. Happy clowning with thermodynamics!
To calculate the maximum amount of work that can be obtained from the combustion of 1.00 mole of ethane (C2H6) at 25 °C and standard conditions, you can use the following steps:
1. First, calculate the standard Gibbs free energy change of the reaction for the combustion of ethane (ΔG°rxn) using the equation: ΔG°rxn = ΣΔG°f(products) - ΣΔG°f(reactants).
2. The ΔG°f values represent the standard Gibbs free energy of formation for each compound. However, since you mentioned that you could not find the value for ΔG°f of ethane (C2H6) in the given table, we will need to find an alternative approach.
3. You can use the bond dissociation energy method to estimate the ΔG°f of ethane. Ethane (C2H6) can be considered as a combination of two carbon-hydrogen (C-H) bonds and six carbon-carbon (C-C) bonds.
4. The bond dissociation energy represents the energy required to break a particular type of bond. The average bond dissociation energies for C-H and C-C bonds are approximately 413 kJ/mol and 348 kJ/mol, respectively.
5. Hence, for ethane (C2H6), the bond dissociation energy would be:
2(C-H bonds) + 6(C-C bonds) = 2(413 kJ/mol) + 6(348 kJ/mol) = 2438 kJ/mol.
6. The ΔG°f for the reactant (ethane) can be calculated using the equation: ΔG°f(reactant) = ΣΔG°f(products) - ΣΔG°f(reactants).
Since ethane is a reactant, it does not have any preceding products, so ΔG°f(reactant) = 0 kJ/mol.
7. Now, we can calculate ΔG°rxn by subtracting the ΔG°f of the reactant from the ΔG°f of the products:
ΔG°rxn = ΔG°f(products) - ΔG°f(reactant) = -84.0 kJ/mol - 0 kJ/mol = -84.0 kJ/mol.
8. Finally, the maximum amount of work obtained from the combustion reaction can be calculated using the equation: maximum work = -ΔG°rxn.
Therefore, the maximum amount of work obtained from the combustion of 1.00 mole of ethane at 25 °C and standard conditions is 84.0 kJ/mol.
To calculate the maximum amount of work that can be obtained from the combustion of 1.00 mole of ethane (CH3CH3), you need to use thermodynamic data such as standard enthalpy of formation (∆Hf), standard Gibbs free energy of formation (∆Gf), and known equations.
First, let's address the information in the hint. The value ∆Gf represents the standard Gibbs free energy of formation. In this case, it refers to the Gibbs free energy change for the formation of one mole of ethane at standard conditions. You mentioned that the value is -30.2 kJ/mol.
Since you are dealing with the combustion of ethane, you will also need to use the standard enthalpy of formation of ethane (∆Hf). However, you did not provide this value in your question, so we will have to make an assumption. Let's assume that the value is -84.0 kJ/mol, as you mentioned it in the table.
To calculate the maximum amount of work (∆W) that can be obtained from the combustion reaction, you can use the equation:
∆G = -∆W (since work is negative when it's being done ON the system)
Then, you can use the following equation to relate ∆G to ∆H and ∆S:
∆G = ∆H - T∆S
Where:
∆G = Gibbs free energy change
∆H = enthalpy change
T = temperature in Kelvin
∆S = entropy change
Since the reaction is at standard conditions, you can assume that the temperature is 298K (25 °C = 298K). Also, at standard conditions, the Gibbs free energy change is equal to the standard Gibbs free energy of formation (∆Gf). Therefore, you can rewrite the equation as:
∆Gf = ∆Hf - T∆S
Now, let's solve for ∆S. Assume that entropy change (∆S) for the formation of ethane (∆Sf) is known and provided in the table. You can use it to find the ∆S value for the reaction:
∆S = ∆Sproducts - ∆Sreactants
In this case, you can calculate ∆S as follows:
∆S = ∆Sf(CH3CH3) - [2∆S(C) + 3∆S(H2)]
Once you have the values of ∆Hf, ∆Gf, and ∆S, you can calculate ∆G using the equation:
∆G = ∆Gf - T∆S
Plug in the values and calculate ∆G. Then, since ∆G = -∆W, the maximum work obtained from the combustion reaction would be the negative value of ∆G.
Remember to ensure that all the values you use are in consistent units (e.g., kJ/mol, kJ/K, etc.) to yield the correct result.
Note: It's important to consult the given table to obtain the correct values of ∆Hf, ∆Sf, and other relevant thermodynamic properties for ethane.
2C2H6 + 7O2 --> 4CO2 + 6H2O
dGrxn = (n*dGo products) - (n*dGo reactants)