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, CH2O(g), which poses a health hazard. Calculate ΔG o at 2.0 × 102°C for this oxidation. × 10 kJ Enter your answer in scientific notation.

2.91x10¹ kJ

Explanation:

First, let's write the chemical equation again:

CO(g) + 2H₂(g) <> CH₃OH(l)

To do this, we should use the following expression:

ΔG° = ΔH° - TΔS°   (1)

Where:

ΔH°: enthalpy of reaction

ΔS°: Entropy of reaction

To get these values, we need the values of ΔH and ΔS of formation of all the reactants and products. You can see these values in any handbook.

For the case of ΔH°:

ΔH° CH₃OH = -238.6 kJ/mol

ΔH° H₂ = 0 kJ

ΔH° CO = -110.5 kJ/mol

For the case of ΔS°:

ΔS° CH₃OH = 126.8 J/mol K

ΔS° H₂ = 130.7 J/mol K

ΔS° CO = 197.7 J/mol K

Now that we have the values, we should use two more expressions to get the ΔH° and ΔS°:

ΔH° = ΔH°f(products) - ΔHf(reactants)   (2)

ΔS° = ΔS°f(products) - ΔS°f(reactants)   (3)

To get this values, we need to take account the number of moles reacting and producing. In this case we have 1 mole of CO and CH₃OH and 2 moles of H₂. Noting this, the ΔH° and ΔS° are:

ΔH° = (-238.6) - (-110.5 + 0) = -128.1 kJ

ΔS° = (126.8) - (197.7 + 2*130.7) = -332.3 J ---> -0.3323 kJ

Now that we have these values, we can use expression (1) to get the ΔG°. The temperature to use is 200 °C or 473 K so:

ΔG° = (-128.1) - 473*(-0.3323)

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