The value of Δ H-Δ E for the following reaction at 27° C will be 2NH3(g) → N2(g) + 3H2(g)

Question

The value of Δ H-Δ E for the following reaction at 27° C will be 2NH3(g) → N2(g) + 3H2(g) (A)  8.314× 273× (-2)  (B)  8.314× 300× (-2)  (C)  8.314× 27× (-2)  (D)  8.314× 300× (2)

Tardigrade Admin · Accepted Answer

2 NH 3 leftharpoons N 2+3 H 2 ∵ Δ H=Δ E+Δ ng R T ∴ Δ H-Δ E=Δ ng R T Δ ng= gaseous product - gaseous reactant =4-2=+2 At 27° C, Δ H-Δ E=2 R T=2 × 8.314 × 300

Q. The value of $Δ H - Δ E$ for the following reaction at $2 7^{\circ} C$ will be
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$

$8.314 \times 273 \times (- 2)$

$8.314 \times 300 \times (- 2)$

$8.314 \times 27 \times (- 2)$

$8.314 \times 300 \times (2)$

$2 N H_{3} ⇌ N_{2} + 3 H_{2}$
$∵ Δ H = Δ E + Δ n_{g} RT$
$∴ Δ H - Δ E = Δ n_{g} RT$
$Δ n_{g} =$ gaseous product $-$ gaseous reactant
$= 4 - 2 = + 2$
At $2 7^{\circ} C$ ,
$Δ H - Δ E = 2 RT = 2 \times 8.314 \times 300$

Q. The value of ΔH−ΔE for the following reaction at 27∘C will be 2NH3​(g)→N2​(g)+3H2​(g)

8.314×273×(−2)

8.314×300×(−2)

8.314×27×(−2)

8.314×300×(2)