½O2(g) + H2(g) ⇌ H2O(g)

Calculate the ΔH, ΔS and ΔSuniverse for this reaction.

Determine the spontaneity of the reaction.

-241.826 kJ·mol⁻¹;  -146.9 J·K⁻¹mol⁻¹; 664.6 J·K⁻¹mol⁻¹; spontaneous

Explanation:

                        ½O₂(g)   +  H₂(g) ⟶ H₂O(g)

ΔHf°/kJ·mol⁻¹:      0                0        -241.826

S°/J·K⁻¹mol⁻¹:   205.0         130.6       188.7

1. ΔᵣH

ΔᵣH = products -reactants = -241.826 -(0 + 0) = -241.826 kJ·mol⁻¹

2. ΔᵣS

ΔᵣS = products - reactants = 188.7 - (205.0 + 130.6) = 188.7 - 335.6 = -146.9 J·K⁻¹mol⁻¹

3. ΔS(univ)

4. Spontaneity

ΔG is negative, so the reaction is spontaneous.

the dependant variable is spilled water

the independent is the heat

the temperature of the gas, in k, not celsius.