Methanol, a major industrial feedstock, is made by several catalyzed reactions, such as CO(g) + 2H2(g) ----> CH3OH (l)
One concern about using CH3OH as an auto fuel is its oxidation in air to yield formaldehyde, CH2OH(g) which poses a health hazard.
Calculate ÄG o at 130 o C for this oxidation (answer must be in scientific notation)
Is it 5.696 KJ ?
I don't know what values you are using. Tables vary so I can't get the same answer unless you give me the numbers you used.
To calculate ΔG° at a given temperature, you'll need to first determine the standard Gibbs free energy change, ΔG°, for the reaction. You can then use the equation ΔG = ΔG° + RTln(Q) to calculate the ΔG at the specified temperature (in this case, 130 °C).
Here's how you can find the standard ΔG° for the oxidation reaction of methanol to formaldehyde:
1. Write the balanced chemical equation for the reaction:
CO(g) + 2H2(g) → CH3OH(l) + ΔG°
2. Find the standard Gibbs free energy change, ΔG°, for each component of the reaction using standard Gibbs free energy of formation values (ΔG°f). The ΔG°f values can usually be found in thermodynamics textbooks or online databases. If we assume that the ΔG°f for CO(g), H2(g), and CH3OH(l) are known, we can use those values to calculate the standard ΔG° for the reaction.
3. Substitute the ΔG°f values into the equation:
ΔG° = ΣnΔG°f(products) - ΣmΔG°f(reactants)
4. Calculate ΔG° using the given values.
Once you have the ΔG°, you can apply the equation ΔG = ΔG° + RTln(Q) to calculate the ΔG at 130 °C. However, to proceed further and verify whether the answer is 5.696 kJ or not, I would need the ΔG° values (in kJ/mol) for each species involved in the reaction and the gas constant R (in J/(mol·K)).
To calculate ΔG° for the oxidation of methanol to formaldehyde, we can use the equation:
ΔG° = ΔH° – TΔS°
Where:
ΔH° = standard enthalpy change
ΔS° = standard entropy change
T = temperature in Kelvin
The standard enthalpy change (ΔH°) for the reaction can be determined using thermodynamic data. The standard entropy change (ΔS°) can be calculated using the equation:
ΔS° = ΣnS°(products) - ΣmS°(reactants)
where:
n = coefficients of products in the balanced equation
m = coefficients of reactants in the balanced equation
S° = standard entropy of each substance
Let's calculate ΔH° and ΔS° for the reaction:
ΔH° = (ΔHf°(CH2OH(g)) - ΔHf°(CH3OH(l)))
The standard enthalpy of formation (ΔHf°) for each substance can be found in thermodynamic tables. For this case:
ΔHf°(CH2OH(g)) = 31.0 kJ/mol
ΔHf°(CH3OH(l)) = -201.2 kJ/mol
ΔH° = (31.0 kJ/mol - (-201.2 kJ/mol)) = 232.2 kJ/mol
Now let's calculate ΔS°:
ΔS° = (ΣnS°(products) - ΣmS°(reactants))
For CH2OH(g):
S°(CH2OH(g)) = 251.7 J/mol·K
n = 1
For CH3OH(l):
S°(CH3OH(l)) = 126.8 J/mol·K
m = 1
ΔS° = (1 * 251.7 J/mol·K) - (1 * 126.8 J/mol·K) = 124.9 J/mol·K
Now we can calculate ΔG° using the equation:
ΔG° = ΔH° – TΔS°
T = 130 + 273.15 K = 403.15 K
ΔG° = (232.2 kJ/mol) - (403.15 K * 0.1249 kJ/mol·K) = 182.64 kJ/mol
Thus, the value for ΔG° at 130°C for this oxidation reaction is 1.8264 x 10^5 J/mol or 182.64 kJ/mol.