For which of the following equations is the change in enthalpy at 25 celcius and 1 atm pressure equal to δhf of ch3oh (l)? a.) ch3oh + 3/2 o2 > co2 + 2h2o (l)b.) ch3oh + 3/2 o2 > co2 + 2h2o (g)c.) 2ch3oh + 3o2 > 2co2 + 4h2o (l)d.) c + 2h2 + 1/2 o2 > ch3oh

D. C + 2H₂ + ¹/₂O₂ → CH₃OH

Explanation:

The standard enthalpy of formation -ΔHf- is defined as the enthalpy change for the formation of 1 mol of a compound from its component elements.

Component elements are any element in its standard state.

For CH₃OH(l) at 25°C and 1 atm of pressure, the component elements are C(s), H₂(g) and O₂(g)

Thus, the equation that represent the standard enthalpy of formation of CH₃OH(l) is:

D. C + 2H₂ + ¹/₂O₂ → CH₃OH

I hope it helps!

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.