The thermochemical equation for the combustion of methanol is given below.

The thermochemical equation for the combustion of methanol is given below.

The Correct Answer and Explanation is:

The thermochemical equation for the combustion of methanol is:

CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(g)
ΔH° = -638.7 kJ/mol

We are asked to find the enthalpy change for the combustion of 8.59 g of CH3OH.

Step-by-step solution:

1. Determine the molar mass of CH3OH:
   • Carbon (C): 12.01 g/mol
   • Hydrogen (H): 1.008 g/mol (x4 for the 4 hydrogens)
   • Oxygen (O): 16.00 g/mol
   Molar mass of CH3OH = 12.01 + (4 x 1.008) + 16.00 = 32.04 g/mol
2. Convert the mass of CH3OH to moles:
   We are given 8.59 g of CH3OH. To convert this to moles: moles of CH3OH=8.59 g32.04 g/mol=0.268 mol\text{moles of CH3OH} = \frac{8.59 \, \text{g}}{32.04 \, \text{g/mol}} = 0.268 \, \text{mol}moles of CH3OH=32.04g/mol8.59g​=0.268mol
3. Calculate the enthalpy change:
   From the equation, we know that 1 mole of CH3OH releases 638.7 kJ of energy when it combusts. So, for 0.268 moles of CH3OH: enthalpy change=0.268 mol×(−638.7 kJ/mol)=−171.3 kJ\text{enthalpy change} = 0.268 \, \text{mol} \times (-638.7 \, \text{kJ/mol}) = -171.3 \, \text{kJ}enthalpy change=0.268mol×(−638.7kJ/mol)=−171.3kJ

Final answer:

The enthalpy change for the combustion of 8.59 g of CH3OH is -171.3 kJ.