Question

3
12. Correct expression of enthalpy of combustion is
[NCERT Pg. 176]
(1) C2H6(g) + O2(g) → 2CO(g) + 3H20(1)
(2) C(s) +-02 (9) --CO(9)
2
(diamond)
4
(3) C(s) +-02 (9)--CO (9)
graphite
2
(4) S(s) + O2(g) →S02 (9)
rhombic​

Answer 1

Answer:

(4)

S+O2——— SO2(g)

Explanation:

1:-in first option is not Balanced ,

2:-carbon (diamond ) form CO2 so

Answer 2

Answer:

Correct expression of enthalpy of combustion is

$\mathrm{\underset{Graphite}{C(s)}+O_2(g)\rightarrow {CO_2(g)}}$     and $\mathrm{\underset{Rhombic}{S(s)}+O_2(g)\rightarrow {SO_2(g)}}$ i.e.option(3) and (4).

Explanation:

First, understand what is the enthalpy of combustion.

• The enthalpy change that occurs when one mole of a substance is entirely burned in oxygen with all of the reactants and products in their standard states under standard conditions is known as the "standard enthalpy of combustion" (298K and 1 bar pressure).
• The standard state of carbon is graphite and that of sulfur is rhombic.

And the reaction proceeds as follows:-

$\mathrm{\underset{Graphite}{C(s)}+O_2(g)\rightarrow {CO_2(g)}}$                         (1)

$\mathrm{\underset{Rhombic}{S(s)}+O_2(g)\rightarrow {SO_2(g)}}$                        (2)

Where,

C(s)=carbon

O₂=oxygen

CO₂=carbon dioxide

SO₂=sulfur dioxide

Hence, the correct answer among the given option is (3) and (4).

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