Given that N 2(g)+3 H 2(g) longrightarrow 2 NH 3(g) Δr H°=-92 kJ, the standard molar enthalpy of formation in kJ mol -1 of NH 3(g) is

Question

Given that N 2(g)+3 H 2(g) longrightarrow 2 NH 3(g) Δr H°=-92 kJ, the standard molar enthalpy of formation in kJ mol -1 of NH 3(g) is (A) -92 (B) +46 (C) +92 (D) -46

Tardigrade Admin · Accepted Answer

Given, N 2(g)+3 H 2(g) longrightarrow 2 NH 3(g) ; Δr H°=-92 kJ Chemical reaction for molar enthalpy of formation of NH 3, (1/2) N 2(g)+(3/2) H 2(g) longrightarrow NH 3(g) (∵ Hf°. for N2 and .H2=0) Therefore, Δf H°=(-Δr H°/2)=(-92/2) =-46 kJ mol -1

Q. Given that N2​(g)+3H2​(g)⟶2NH3​(g) Δr​H∘=−92kJ, the standard molar enthalpy of formation in kJmol−1 of NH3​(g) is

-92

+46

+92

-46

Given, N2​(g)+3H2​(g)⟶2NH3​(g);Δr​H∘=−92kJ Chemical reaction for molar enthalpy of formation of NH3​, 21​N2​(g)+23​H2​(g)⟶NH3​(g) (∵Hf∘​ for N2​ and H2​=0) Therefore, Δf​H∘=2−Δr​H∘​=2−92​ =−46kJmol−1

Q. Given that $N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} (g)$ $Δ_{r} H^{\circ} = - 92 k J$ , the standard molar enthalpy of formation in $k J m o l^{- 1}$ of $N H_{3} (g)$ is