Given N2(g)+3H2(g)=2NH3(g);Δ H°=-22 kcal.. The standard enthalpy of formation of NH3 gas is

Question

Given N2(g)+3H2(g)=2NH3(g);Δ H°=-22 kcal.. The standard enthalpy of formation of NH3 gas is (A) - 11 kcal/mol (B) 11 kcal/mol (C) - 22 kcal/mol (D) 22 kcal/mol

Tardigrade Admin · Accepted Answer

Δ H°= Sigma nΔ H°f( textproduct)- Sigma nΔ H°f( textreactant) =2Δ H°f(NH3)-(Δ H°f(N2)+3Δ H°f(H2),)Δ H°=22 kcal =Δ H°f(NH3)-(0+0) So, Δ H°f=-11 kcal

Q. Given
$N_{2} (g) + 3 H_{2} (g) = 2 N H_{3} (g); Δ H^{\circ} = - 22 k c a l .$ . The standard enthalpy of formation of $N H_{3}$ gas is

$- 11 k c a l / m o l$

$11 k c a l / m o l$

$- 22 k c a l / m o l$

$22 k c a l / m o l$

$Δ H^{^{\circ}} = Σ n Δ H_{f}^{^{\circ}} (product) - Σ n Δ H_{f}^{^{\circ}} (reactant)$ $= 2Δ H_{f}^{\circ} (N H_{3}) - (Δ H_{f}^{\circ} (N_{2}) + 3Δ H_{f}^{\circ} (H_{2}),) Δ H^{\circ} = 22$ kcal
$= Δ H_{f}^{\circ} (N H_{3}) - (0 + 0)$
So, $Δ H_{f}^{\circ} = - 11$ kcal

Q. Given N2​(g)+3H2​(g)=2NH3​(g);ΔH∘=−22kcal.. The standard enthalpy of formation of NH3​ gas is

−11kcal/mol

11kcal/mol

−22kcal/mol

22kcal/mol