question archive For which of the following reactions does ΔHorxn = ΔHof? (a) H2(g) + S(rhombic) → H2S(g) (b) C(diamond) + O2(g) → CO2(g) (c) H2(g) + Cu O(s) → H2O(l) + Cu(s) (d) O(g) + O2(g) → O3(g)
For which of the following reactions does ΔHorxn = ΔHof? (a) H2(g) + S(rhombic) → H2S(g) (b) C(diamond) + O2(g) → CO2(g) (c) H2(g) + Cu O(s) → H2O(l) + Cu(s) (d) O(g) + O2(g) → O3(g)
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For which of the following reactions does ΔHorxn = ΔHof?
(a) H2(g) + S(rhombic) → H2S(g)
(b) C(diamond) + O2(g) → CO2(g)
(c) H2(g) + Cu O(s) → H2O(l) + Cu(s)
(d) O(g) + O2(g) → O3(g)
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The reaction where ΔHorxn = ΔHof is H2 (g) + S (rhombic) --------> H2S (g)
Therefore, option (a) is the correct one.
Step-by-step explanation
Among the reactions given above, ΔHorxn = ΔHof is true for only one reaction. That is reaction (a)
where H2 (g) + S (rhombic) --------> H2S (g)
For the standard heat enthalpy of a reaction (ΔHorxn ) to be equal to the standard heat of formation (ΔHof ), the compound formed in the product side is formed from its constituent elements in their standard states. Looking at all the reaction given, only reaction (a) meets this criteria.
For reaction (b), the standard state of C is graphite and not diamond as indicated in this equation. Hence C is not in its standard state.
For reaction (c) and reaction (d), some of the reactants are not in their standard states. Hence their heat enthalpy of a reaction (ΔHorxn ) can't be equal to standard heat of formation (ΔHof ).
Therefore, it is only reaction (a) : H2 (g) + S (rhombic) --------> H2S (g) where ΔHorxn = ΔHof
