The enthalpy of formation of ammonia is -46.0 kJ mol -1 . The enthalpy for the reaction 2 N 2(g)+6 H 2(g) arrow 4 NH 3( g ) is equal to -

Question

The enthalpy of formation of ammonia is -46.0 kJ mol -1 . The enthalpy for the reaction 2 N 2(g)+6 H 2(g) arrow 4 NH 3( g ) is equal to - (A) -46.0 kJ (B) 46.0 kJ (C) 184.0 kJ (D) -184.0 kJ

Tardigrade Admin · Accepted Answer

2 N 2( g )+6 H 2( g ) arrow 4 NH 3( g ) Δ H =-46 kJ / mole Formation of 1 . mole NH 4=46 kJ / mole 4 mole NH 4=-4 × 4.6 =-184 kJ / mole

Q. The enthalpy of formation of ammonia is $- 46.0 k J$ $m o l^{- 1} .$ The enthalpy for the reaction $2 N_{2} (g) + 6 H_{2} (g) \to$ $4 N H_{3} (g)$ is equal to -

-46.0 kJ

46.0 kJ

184.0 kJ

-184.0 kJ

$2 N_{2} (g) + 6 H_{2} (g) \to 4 N H_{3} (g)$

$Δ H = - 46 k J / m o l e$

Formation of $1 . m o l e N H_{4} = 46 k J / m o l e$

4 mole $N H_{4} = - 4 \times 4.6$

$= - 184 k J / m o l e$

Q. The enthalpy of formation of ammonia is −46.0kJ mol−1. The enthalpy for the reaction 2N2​(g)+6H2​(g)→ 4NH3​(g) is equal to -