C 6 H 12(l)+9 O 2( g ) arrow 6 H 2 O (l)+6 CO 2( g ) ; Δ H 298=-936.9 kcal. Thus

Question

C 6 H 12(l)+9 O 2( g ) arrow 6 H 2 O (l)+6 CO 2( g ) ; Δ H 298=-936.9 kcal. Thus (A) -936.9=Δ E -(2 × 10-3 × 298 × 3) kcal (B) +936.9=Δ E +(2 × 10-3 × 298 × 3) kcal (C) -936.9=Δ E -(2 × 10-3 × 298 × 2) kcal (D) -936.9=Δ E +(2 × 10-3 × 298 × 2) kcal

Tardigrade Admin · Accepted Answer

Δ ng=6-9=-3 ∴ Δ H=Δ U+Δ ng R T or, Δ U=Δ H-Δ ng R T =-936.9 kCal =(-3 R × 298 K ) or, -936.9 kcal =Δ U+(-3 × 2 cal k-1 m o l-1 × 298 K ) =Δ U-(3 × 2 × 298 × 10-3) kcal Δ U =Δ E = change in internal energy

Q. $C_{6} H_{12} (l) + 9 O_{2} (g) \to 6 H_{2} O (l) + 6 C O_{2} (g); Δ H_{298} = - 936.9 k c a l$ . Thus

$- 936.9 = Δ E - (2 \times 1 0^{- 3} \times 298 \times 3) k c a l$

$+ 936.9 = Δ E + (2 \times 1 0^{- 3} \times 298 \times 3) k c a l$

$- 936.9 = Δ E - (2 \times 1 0^{- 3} \times 298 \times 2) k c a l$

$- 936.9 = Δ E + (2 \times 1 0^{- 3} \times 298 \times 2) k c a l$

Q. C6​H12​(l)+9O2​(g)→6H2​O(l)+6CO2​(g);ΔH298​=−936.9kcal. Thus

−936.9=ΔE−(2×10−3×298×3)kcal

+936.9=ΔE+(2×10−3×298×3)kcal

−936.9=ΔE−(2×10−3×298×2)kcal

−936.9=ΔE+(2×10−3×298×2)kcal

Δng​=6−9=−3 ∴ΔH=ΔU+Δng​RT or, ΔU=ΔH−Δng​RT =−936.9kCal=(−3R×298K) or, −936.9kcal=ΔU+(−3×2calk−1mol−1×298K) =ΔU−(3×2×298×10−3)kcal ΔU=ΔE= change in internal energy

Q. $C_{6} H_{12} (l) + 9 O_{2} (g) \to 6 H_{2} O (l) + 6 C O_{2} (g); Δ H_{298} = - 936.9 k c a l$ . Thus

$- 936.9 = Δ E - (2 \times 1 0^{- 3} \times 298 \times 3) k c a l$

$+ 936.9 = Δ E + (2 \times 1 0^{- 3} \times 298 \times 3) k c a l$

$- 936.9 = Δ E - (2 \times 1 0^{- 3} \times 298 \times 2) k c a l$

$- 936.9 = Δ E + (2 \times 1 0^{- 3} \times 298 \times 2) k c a l$