How would you exactly start the first problem? My thermodynamic skills is a bit rusty. I am assuming I have to use ΔH°f under the given reaction..? But then..we barely covered ΔG° in class. Am I to use this equation, ΔG° = ΔH° - T ΔS°? Umm..can someone explain in a detailed process how to go about this problem? Also for problem #2, I don't really understand what the question is asking for.
1. Consider the following reaction and thermodynamic data:
Sr(s) + H2O(g) --> SrO(s) + H2(g)
Substance________ΔH°f (/kJ/mol)________S° (J/mol K)
Sr(s)____0____52.3
H2O(g)____-241.8____188.7
SrO(s)____-634.9____54.4
H2(g)____0____130.6
A) What is ΔG° at 500.0 K for the above reaction?
B) Calculate the ΔG at 500.0 K if the reaction is run with PH2O=5.0 atm and PH2=1.0 atm.
2. An ideal fuel for the jet of a space vehicle should decompose with
A) ΔG < 0 and ΔH < 0
B) ΔG < 0 and ΔH > 0
C) ΔG > 0 and ΔH < 0
D) ΔG > 0 and ΔH > 0
E) ΔG = 0 and ΔH = 0
Yes, for #1 use dGo = dHo - TdSo. That will give you dGo. Then if you assume dH = dHo and dS = dSo and use T = 500K.
I think for #2 you are to use Keq expression with dG which is dG = -RTlnK. and go from there. So you know R, and T,
K = p(H2)/p(H2O) substitute and solve for dG.
What I stated as answer for #2 actually is B of #1.
Thank you DrBob22. :) But for #1, am I looking for dGo for H2(g)?
Or calculate dHo and dSo at standard enthalpies of formation and entropies?
To start solving the first problem, you can determine the ΔG° at 500.0 K for the reaction using the equation ΔG° = ΔH° - TΔS°. Here's a step-by-step process to solve the problem:
1. Identify the substances involved in the reaction and their corresponding ΔH°f and S° values:
- Sr(s): ΔH°f = 0 kJ/mol, S° = 52.3 J/mol K
- H2O(g): ΔH°f = -241.8 kJ/mol, S° = 188.7 J/mol K
- SrO(s): ΔH°f = -634.9 kJ/mol, S° = 54.4 J/mol K
- H2(g): ΔH°f = 0 kJ/mol, S° = 130.6 J/mol K
2. Write down the balanced equation for the reaction:
Sr(s) + H2O(g) → SrO(s) + H2(g)
3. Calculate the ΔH° for the reaction by summing the ΔH°f values of the products and subtracting the sum of the ΔH°f values of the reactants:
ΔH° = Σ(ΔH°f products) - Σ(ΔH°f reactants)
= [ΔH°f(SrO) + ΔH°f(H2)] - [ΔH°f(Sr) + ΔH°f(H2O)]
= [-634.9 + 0] - [0 + (-241.8)]
= -634.9 + 241.8
= -393.1 kJ/mol
4. Calculate the ΔS° for the reaction by summing the S° values of the products and subtracting the sum of the S° values of the reactants:
ΔS° = Σ(S° products) - Σ(S° reactants)
= [S°(SrO) + S°(H2)] - [S°(Sr) + S°(H2O)]
= [54.4 + 130.6] - [52.3 + 188.7]
= 185 - 241
= -56 J/mol K
5. Substitute the calculated ΔH° and ΔS° values into the equation ΔG° = ΔH° - TΔS°:
ΔG° = ΔH° - TΔS°
= -393.1 - (500 * -56)
= -393.1 + 28000
= 27606.9 J/mol
Since ΔG° is in J/mol, you can convert it to kJ/mol by dividing by 1000:
ΔG° = 27.6069 kJ/mol
For part B, you need to calculate ΔG at 500.0 K if the reaction is run under specific pressure conditions. The equation to calculate ΔG at non-standard conditions is:
ΔG = ΔG° + RTln(Q)
Where:
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
- Q is the reaction quotient
However, since the reaction given in part B involves gases, we can assume ideal gas behavior and use the equation:
ΔG = ΔG° + RTln(Q/P°)
Where P° is the standard pressure (1 atm).
2. For the second problem, the question is asking for the ideal fuel for a space vehicle. An ideal fuel for a jet should decompose, which means the reaction is exothermic and releases energy. Therefore, the correct answer would be:
A) ΔG < 0 and ΔH < 0
This means that the change in Gibbs free energy (ΔG) and the change in enthalpy (ΔH) both need to be negative for the ideal fuel.