For the reaction CH3OH(g)←→CO(g) + 2H2(g),

a. Determine ΔHo and ΔSo for the reaction at 298K
b. Determine the entropy change for the surroundings assuming that they are at 298 K. What does this result tell you?
c. Calculate ΔGo at 298 from tabulated ΔG𝑓𝑜 values. Is it consistent with your result from b?
d. Assuming that ΔHo and ΔSoremain relatively constant at different temperatures, calculate ΔGoat 28 oC and 228 oC.
e. From you answer from c, would it be wiser to conduct the experiment at higher or lower temperatures if you wished the reaction to produce hydrogen gas.
f. At what temperature is the reaction at equilibrium?

Long question. How much do you already know how to do? Exactly what do you not understand. Regarding a part, what keeps you from looking up the values for dHo and dSo? Same for dGo for c.

I pretty much understand nothing

That doesn't help much. We need to know what to focus on to provide the best meaningful help.

So look up the values of dHo and dSo.
dHorxn = (n*dH formation products) - (n*dHo formation reactants.
dSorxn = (n*dSo formation products) - (n*dSo formation reactants)

dSsurr = -dH/T

Look up dGo and calculate dGorxn the same way you did dHorxn and dSorxn above.

d. dG = dH - TdS

e. Use the information above.

f. Set dG = 0 and solve for T.
Post your work if you get stuck and PLEASE explain what you don't understand.

a. To determine ΔHo and ΔSo for the reaction at 298K, we can use the standard enthalpy of formation (ΔH𝕗𝕠) and standard entropy (S𝕠) values for the reactants and products. ΔHo is the difference in enthalpy between the products and reactants, and ΔSo is the difference in entropy between the products and reactants.

The standard enthalpy change (ΔHo) can be calculated using the equation:
ΔHo = Σ(nΔH𝕗𝕠(products)) - Σ(mΔH𝕗𝕠(reactants))

The standard entropy change (ΔSo) can be calculated using the equation:
ΔSo = Σ(nS𝕠(products)) - Σ(mS𝕠(reactants))

Where n and m are the stoichiometric coefficients of the products and reactants, ΔH𝕗𝕠 is the standard enthalpy of formation, and S𝕠 is the standard entropy.

b. To determine the entropy change for the surroundings, we can use the equation:
ΔSsurr = -ΔHo/T

Where ΔSsurr is the entropy change of the surroundings, ΔHo is the standard enthalpy change, and T is the temperature in Kelvin. This equation assumes that the reaction occurs at constant pressure and that the only work done is by the system on the surroundings.

This result tells us that the entropy of the surroundings decreases when the reaction proceeds in the forward direction (from left to right). This is because the formation of the products releases heat, causing an increase in the entropy of the system (products) and a decrease in the entropy of the surroundings (since heat is leaving the surroundings).

c. To calculate ΔGo at 298 K, we can use the equation:
ΔGo = Σ(nΔG𝕗𝕠(products)) - Σ(mΔG𝕗𝕠(reactants))

Where ΔG𝕗𝕠 is the standard free energy of formation. We can obtain the values of ΔG𝕗𝕠 from tabulated data.

Comparing the value from part b with the calculated ΔGo, we can determine if they are consistent. If the sign of ΔGo matches the sign of ΔSsurr (negative for an exothermic reaction), then they are consistent.

d. To calculate ΔGo at different temperatures, we can use the equation:
ΔGo = ΔHo - TΔSo

Where ΔHo and ΔSo are the enthalpy and entropy changes calculated in part a, and T is the temperature in Kelvin.

e. From the result from part c, if ΔGo is negative, the reaction is favorable at lower temperatures. This suggests that it would be wiser to conduct the experiment at lower temperatures to produce hydrogen gas.

f. To find the temperature at which the reaction is at equilibrium, we can use the equation:
ΔGo = -RTln(K)

Where ΔGo is the standard free energy change, R is the gas constant (8.314 J/mol*K), T is the temperature in Kelvin, and K is the equilibrium constant. Rearranging the equation, we can solve for T.

This temperature gives the value at which the forward and reverse reactions have equal rates, resulting in equilibrium.