calculate the standard molar entropy change for the combustion of methane gas using S naught values from standard thermodynamic tables. Assume that liquid water is one of the products.
CH4(g) + 2O2(g) ==> CO2(g) + H2O(l)
dSo rxn = (n*dSo poducts) - (dSo reactants)
Thank you!
Oh boy, it's time for some methane combustion fun! Now, before diving into the standard molar entropy change calculations, let me tell you a little methane combustion joke:
Why did the methane gas go to therapy?
Because it had too much internal pressure!
Now, let's get serious and calculate that entropy change. The standard molar entropy change for the combustion of methane gas can be determined using the entropy values provided in standard thermodynamic tables. When methane (CH4) combusts, it reacts with oxygen (O2) to form carbon dioxide (CO2) and water (H2O), as represented by the equation:
CH4 + 2O2 -> CO2 + 2H2O
Now, let's break it down. The standard molar entropy change for the reaction can be calculated as follows:
ΔS° = ΣnS°(products) - ΣmS°(reactants)
Where:
ΔS° is the standard molar entropy change,
ΣnS°(products) is the sum of the standard molar entropies of the products, and
ΣmS°(reactants) is the sum of the standard molar entropies of the reactants.
Looking up the standard molar entropy values in a thermodynamic table, we find:
S°(CH4) = 186.3 J/mol·K
S°(O2) = 205 J/mol·K
S°(CO2) = 213.7 J/mol·K
S°(H2O) = 188.7 J/mol·K
Now plug these values into the equation:
ΔS° = [1 × S°(CO2) + 2 × S°(H2O)] - [1 × S°(CH4) + 2 × S°(O2)]
= [1 × 213.7 + 2 × 188.7] - [1 × 186.3 + 2 × 205]
After doing the math, the standard molar entropy change comes out to be quite boring. Oops, I mean... *drumroll* the value is approximately -37 J/mol·K.
So, in the combustion of methane gas with liquid water as one of the products, there is a decrease in entropy. That's a bit of a bummer, but hey, at least we had some fun along the way, right? Keep the questions coming, and let's keep the laughter going!
To calculate the standard molar entropy change (ΔS°) for the combustion of methane gas with liquid water as one of the products, you will need to use the standard entropy values (S°) from the thermodynamic tables.
The balanced chemical equation for the combustion of methane gas can be written as:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)
The molar coefficients in the balanced equation represent the stoichiometric ratio of the substances involved. So, one mole of methane reacts with two moles of oxygen gas to produce one mole of carbon dioxide gas and two moles of liquid water.
To calculate the standard molar entropy change, you can use the equation:
ΔS° = ΣnS°(products) - ΣmS°(reactants)
Where:
Σn: Represents the coefficient of the products
Σm: Represents the coefficient of the reactants
S°(products): Standard entropy of the products
S°(reactants): Standard entropy of the reactants
Now, let’s calculate the standard molar entropy change for the combustion of methane gas:
- First, find the standard entropy values (S°) from the thermodynamic tables for each compound involved in the reaction.
Methane (CH4): S° = xx J/(mol•K)
Oxygen (O2): S° = xx J/(mol•K)
Carbon dioxide (CO2): S° = xx J/(mol•K)
Water (H2O): S° = xx J/(mol•K)
Note: The values of S° will depend on the thermodynamic tables you are using, and the units will be in J/(mol•K).
- Now, substitute these values into the entropy change equation:
ΔS° = (1 mol) * S°(CO2) + (2 mol) * S°(H2O) - (1 mol) * S°(CH4) - (2 mol) * S°(O2)
- Calculate the values for S° and substitute them into the equation above.
ΔS° = (1 mol) * S°(CO2) + (2 mol) * S°(H2O) - (1 mol) * S°(CH4) - (2 mol) * S°(O2)
- Perform the calculations and obtain the final value for the standard molar entropy change (ΔS°) for the combustion of methane gas using the S° values from the thermodynamic tables.
Keep in mind that the above values are placeholders and you need to refer to the specific thermodynamic tables to find the actual S° values.
I did this and the database said the answer was wrong:
(213.8 + 70.0) - (186.2 + 2*205.0)= -312.4 j/K*mol. Am I missing something?
I'm sorry. I didn't balance the equation. I should have written (and you should have checked)
CH4 + 2O2 ==> CO2 + 2H2O(l)
Try that and see if you have better luck. Thanks for letting me know. I almost always check the equations; this time I didn't and it caught me.