calculate delta G at 298 K when 5.0 atm H2 and 3.0 atm of CO2 are converted to methanol.
I may have stared at the sun too long today but why isn't dG = 0 when the reaction has reached equilibrium? You can calculate those concentrations if you wish AND you can calculate Kp at 25C with dGo = -RTlnK. But I don't think any of that is necessary since we know dG = 0 at equilibrium. And the problems says WHEN CO2 and H2 are converted which means AFTER the reaction to me.
.........CO2 + 3H2 ==> CH3OH + H2O
I........ 5.....3........0......0
C........-p....-p........p......p
E........5-p...3-p.......p......p
Calculate p after you know Kp if you really need to know those pressures.
To calculate the change in Gibbs free energy (ΔG) at 298 K for the conversion of H2 and CO2 to methanol, you need to use the equation:
ΔG = ΔG° + RTln(Q)
Where:
ΔG is the change in Gibbs free energy
ΔG° is the standard Gibbs free energy change
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (298 K in this case)
ln is the natural logarithm
Q is the reaction quotient
First, you need to write the balanced chemical equation for the reaction:
CO2 + 3H2 ⇌ CH3OH
Next, determine the standard Gibbs free energy change (ΔG°) for the reaction. This value can be found in thermodynamic tables or calculated using the standard enthalpies of formation of the reactants and products.
Let's assume ΔG° for the reaction is -50 kJ/mol (a negative value indicates a spontaneous reaction).
Now, you need to calculate the reaction quotient (Q) using the given partial pressures of H2 and CO2:
Q = (PH2)^a × (PCO2)^b
Where:
PH2 is the partial pressure of H2 (5.0 atm)
PCO2 is the partial pressure of CO2 (3.0 atm)
a is the stoichiometric coefficient of H2 (-3 in this case)
b is the stoichiometric coefficient of CO2 (-1 in this case)
Calculating Q:
Q = (5.0 atm)^-3 × (3.0 atm)^-1
Now, substitute the values into the equation:
ΔG = -50,000 J/mol + (8.314 J/(mol·K)) × (298 K) × ln[(5.0 atm)^-3 × (3.0 atm)^-1]
Simplify and calculate ΔG to get the final answer.