1. Methanol is a high-octane fuel used in high performance racing engines. 2 CH3OH(l) + 3O2(g) → 2CO2(g) + 4 H20(g) a) Calculate ∆H० and ∆S० using thermodynamic data, and then ∆G

The reaction given in the question is:  

2CH₃OH (l) + 3O₂ (g) ⇒ 2CO₂ (g) + 4H₂O (g)

The values of ΔH°formation and ΔS° of the reactants and products given in the reaction based on the thermodynamics data is:

ΔH°formation values of CH3OH (l) is -238.4 kJ/mol, CO2(g) is -393.52 kJ/mol, H2O (g) is -241.83 kJ/mol and O2 (g) is 0.  

The S° values of CH3OH (l) is 127.19 J/molK, CO2(g) is 213.79 J/molK, H2O (g) is 188.84 J/moleK, and O2 (g) is 205.15 J/molK.  

Now the values of ΔH° and ΔS° are,  

ΔH°rxn = 2 * ΔH°formation CO2 (g) + 4 * ΔH°formation H2O (g) - 2*ΔH°formation CH3OH (l)

ΔH°rxn = 2 * (-393.52) + 4 (-241.83) -2 * (-238.4)

ΔH°rxn = -1277.56 kJ/mole

ΔS°rxn = 2 * S° CO2 (g) + 4 * S° H2O (g) - 2*S° CH3OH (l) - 3 * S° O2 (g)

ΔS°rxn = 2 * 213.79 + 4 * 188.84 - 2 * 127.19 - 3*205.15

ΔS°rxn = 313.11 J/mole/K

Now the formula for calculating ΔG°rxn is,  

ΔG°rxn = ΔH°rxn - TΔS°rxn

ΔG°rxn = -1277.56 * 1000 J/mole - 298 * 313.11 J/mole

ΔG°rxn = -1370.86 kJ/mol

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.