For the reaction A(g)2B(g)→ 2C(g)+3D(g), the value of Δ H at 27°C is 19.0 kcal. The value of Δ E for the reaction would be (R = 2.0 cal K-1mol-1)

Question

For the reaction A(g)2B(g)→ 2C(g)+3D(g), the value of Δ H at 27°C is 19.0 kcal. The value of Δ E for the reaction would be (R = 2.0 cal K-1mol-1) (A) 20.8 kcal (B) 19.8 kcal (C) 18.8 kcal (D) 17.8 kcal

Tardigrade Admin · Accepted Answer

Δ H=Δ E+Δ ngRT 19=Δ E+((2×2×300/1000))kcal 19=Δ E+1.2 Δ E=19-1.2 kcal =17.8 kcal

Q. For the reaction
$A (g) 2 B (g) \to 2 C (g) + 3 D (g),$
the value of $Δ H$ at $27° C$ is $19.0 k c a l$ . The value of $Δ E$ for the reaction would be $(R = 2.0 c a l K^{- 1} m o l^{- 1})$

$20.8 k c a l$

$19.8 k c a l$

$18.8 k c a l$

$17.8 k c a l$

$Δ H = Δ E + Δ n_{g} RT$
$19 = Δ E + (\frac{2 \times 2 \times 300}{1000}) k c a l$
$19 = Δ E + 1.2$
$Δ E = 19 - 1.2 k c a l = 17.8 k c a l$

Q. For the reaction A(g)2B(g)→2C(g)+3D(g), the value of ΔH at 27°C is 19.0kcal. The value of ΔE for the reaction would be (R=2.0calK−1mol−1)

20.8kcal

19.8kcal

18.8kcal

17.8kcal