Question

What is the dH of the following rxn? H₂(g) + 1/2 O₂(g) --> H₂O(g)

a) Positive
b) Negative
c) Zero
d) Variable

Answer 1

The enthalpy change
`(ΔH)` for the formation of water vapor
`(H2O(g))` from hydrogen
`(H₂(g))` and oxygen
`(O₂(g))` is negative because the reaction is exothermic.

Step-by-step explanation:

The question asks about the enthalpy change
`(ΔH)` for

the reaction:
`H₂(g) + ½ O₂(g) → H₂O(g)` . The enthalpy change for the formation of water from hydrogen and oxygen gases is typically a negative value. This is because the reaction is exothermic; it releases energy in the form of heat as chemical bonds are formed in the product (water), which are stronger than those in the reactants (hydrogen and oxygen gases).

A known reaction for the formation of water is
`2H₂(g) + O₂(g) → 2H₂O(g)` . The enthalpy of this reaction is negative, indicating that energy is released. Thus for the given reaction, which is essentially half of the formation process, the enthalpy change (ΔH) is also expected to be negative.

When considering the reaction
`H₂O(l) → H₂O(g)` there is an enthalpy change associated with the phase change from liquid to gas, which would also be positive because energy is required to vaporize the water However, this is a separate process and should not be confused with the formation of water vapor from its elements.

Therefore, the correct answer to the given question is b) Negative.