312. Correct expression of enthalpy of combustion is[NCERT Pg. 176](1)

1.	312. 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)graphite2(4) S(s) + O2(g) →S02 (9)rhombic
Answer» Explanation: C 2 H 6 (g)+ 2 7 O 2 (g)→2CO 2 (g)+3H 2 O(L) Here heat of combustion for two mole of C 2 H 6 =3120 kj , So for ONE mole of C 2 H 6 =1560 kj Now we KNOW that ΔH reaction 0 =ΣΔH f 0 Product−ΣΔH f 0 Reactant Now using this formula -1560 = (−2×395−3×285)−X So X = -85 KJ / mol -X = 85

312. 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)graphite2(4) S(s) + O2(g) →S02 (9)rhombic

Answer»

Explanation:

(g)+

(g)→2CO

(g)+3H

O(L)

Here heat of combustion for two mole of C

=3120 kj , So for ONE mole of C

=1560 kj

Now we KNOW that ΔH

reaction

=ΣΔH

Product−ΣΔH

Reactant Now using this formula

-1560 = (−2×395−3×285)−X

So X = -85 KJ / mol

-X = 85