Calculate the standard enthalpy change for the following reaction at 25 °C.?

2CH3 OH(g) + 3O2(g) -------- 2CO2(g) + 4H2O (g)

To calculate the standard enthalpy change for a reaction at 25 °C, you will need to use Hess's Law and the standard enthalpy of formation values for the compounds involved.

Step 1: Write the balanced chemical equation for the reaction.
2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g)

Step 2: Determine the standard enthalpy of formation (∆Hf°) for each compound in the reaction. The standard enthalpy of formation values are usually given in a reference table.

- ∆Hf° for CH3OH(g) = -239.2 kJ/mol
- ∆Hf° for CO2(g) = -393.5 kJ/mol
- ∆Hf° for H2O(g) = -241.8 kJ/mol
- ∆Hf° for O2(g) = 0 kJ/mol (since it is in its standard state)

Step 3: Apply Hess's Law. According to Hess's Law, the standard enthalpy change (∆H°) for a reaction can be calculated by subtracting the sum of the standard enthalpy of formation of the reactants from the sum of the standard enthalpy of formation of the products.

∆H° = (2∆Hf°[CO2] + 4∆Hf°[H2O]) - (2∆Hf°[CH3OH] + 3∆Hf°[O2])

Step 4: Plug in the values and calculate.
∆H° = (2(-393.5 kJ/mol) + 4(-241.8 kJ/mol)) - (2(-239.2 kJ/mol) + 3(0 kJ/mol))
= (-787 kJ/mol) - (-478.4 kJ/mol)
= -308.6 kJ/mol

Therefore, the standard enthalpy change (∆H°) for the given reaction at 25 °C is -308.6 kJ/mol.

dHrxn = (n*dHf products) - (n*dHf reactants)