The enthalpy change (Δ H ) for the reaction, N 2( g )+3 H 2( g ) longrightarrow 2 NH 3( g ) is -92.38 kJ at 298 K. The internal energy change Δ U at 298 K is:

Question

The enthalpy change (Δ H ) for the reaction, N 2( g )+3 H 2( g ) longrightarrow 2 NH 3( g ) is -92.38 kJ at 298 K. The internal energy change Δ U at 298 K is: (A) -92.38 kJ (B) -87.42 kJ (C) -97.34 kJ (D) -89.9 kJ

Tardigrade Admin · Accepted Answer

Δ H =Δ E + P Δ V Δ H =Δ E +Δ nRT Δ E =Δ H -Δ nRT =-92.38-(-2)(8.314) × 10-3 ×(298) Δ E =-92.38+4.895 =-87.48 kJ

Q. The enthalpy change $(Δ H)$ for the reaction, $N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} (g)$ is $- 92.38 k J$ at $298 K$ . The internal energy change $Δ U$ at $298 K$ is:

-92.38 kJ

-87.42 kJ

-97.34 kJ

-89.9 kJ

$Δ H = Δ E + P Δ V$
$Δ H = Δ E + Δ n RT$
$Δ E = Δ H - Δ n RT$
$= - 92.38 - (- 2) (8.314) \times 1 0^{- 3} \times (298)$
$Δ E = - 92.38 + 4.895$
$= - 87.48 k J$

Q. The enthalpy change (ΔH) for the reaction, N2​(g)+3H2​(g)⟶2NH3​(g) is −92.38kJ at 298K. The internal energy change ΔU at 298K is: