Calc ∆Ho & ∆Eo for CH3OH Reaction @ 298K

[Ariel Jo]

Homework Statement

For the reaction CH₃OH(l) → CH₄ (g)+ 1/2 O₂ (g)[/B]
(a) Calculate ∆H^o₂₉₈

(b) Calculate ∆E^o₂₉₈

(c) Write an equation that would allow you to determine ∆H at 500°C and 1atm

Homework Equations

∆H=∆E+∆(PV)
∆H=Σ∆H_products-Σ∆H_reactants

The Attempt at a Solution

a) [∆(CH₄)+.5(∆O₂)]-∆(CH₃OH)=[-74.87 kJ/mol + .5(0 kJ/mol)]-(-238.4 kJ/mol)=∆^o₂₉₈=163.53 kJ/mol

b) I'm lost on how to find the change in internal energy of the system here... Is there a PV integral for work, maybe? If I can't solve it, I may be a smart ass and just write "energy is conserved, so ΔE is 0 for the universe"

c) ΔG^o=ΔH^o-TΔS^o would be my approach here, any feedback on whether this is correct and why would be appreciated

 

[Chestermiller]

Part b):

To get ΔE, you have to subtract Δ(PV). What is the molar volume of methanol liquid? You can use the ideal gas law to get the molar volume of methane and oxygen.

In part c), are you asking about ΔG or ΔH. There's nothing after the Δ.

Chet

 

[Ariel Jo]

Part b):

To get ΔE, you have to subtract Δ(PV). What is the molar volume of methanol liquid? You can use the ideal gas law to get the molar volume of methane and oxygen.

In part c), are you asking about ΔG or ΔH. There's nothing after the Δ.

Chet

Thank you! I will try the enthalpy=energy+pv equation.
in (c) it was meant to be delta H, I've edited it so it should appear now.

 

[Chestermiller]

For part c), you just start with ΔH at 298K and 1 atm, and use molar heat capacities in conjunction with Hess's law to get ΔH at 500 K. Remember Hess's law?

Chet

 

[Ariel Jo]

Yup! The sum of the state functions of the products minus the sum of the state functions of the reactants is equal to the sum of the state function of the reaction; change in enthalpy is a state function, so it applies here.

 

[Chestermiller]

Yup! The sum of the state functions of the products minus the sum of the state functions of the reactants is equal to the sum of the state function of the reaction; change in enthalpy is a state function, so it applies here.

So, you're able to do part (c) now, right?

Chet

 

[Ariel Jo]

Would it look something like:
ΔH₅₀₀=ΔH^o+(500-298)*___kJ/mol*___mol
for each component of the reaction?
Where ___kJ/mol is the molar heat capacity at constant pressure, and ___mol is the quantity of substance

 

[Chestermiller]

Would it look something like:
ΔH₅₀₀=ΔH^o+(500-298)*___kJ/mol*___mol
for each component of the reaction?
Where ___kJ/mol is the molar heat capacity at constant pressure, and ___mol is the quantity of substance

yes, but you would have to cool the reactant down to 298 and heat the products back up to 500.