The enthalpy of formation of ammonia is -46 2kJ mol-1. The enthalpy change for the reaction 2NH3 ----- N2 + 3H2 is

Question

The enthalpy of formation of ammonia is -46 2kJ mol-1. The enthalpy change for the reaction 2NH3 -----> N2 + 3H2 is (A) 42 .0 kJ (B) 64 .0 kJ (C) 80 .0 kJ (D) 92 .0 kJ

Tardigrade Admin · Accepted Answer

2NH3 arrow N2 + 3H2 2 moles The enthalpy of formation of ammonia = - 46 kJ mol-1 ΔH of given reaction = -2 × (- 46) = 92 kJ

Q. The enthalpy of formation of ammonia is $- 462 k J m o l^{- 1}$ . The enthalpy change for the reaction $2 N H_{3} - - - - - > N_{2} + 3 H_{2}$ is

42 .0 kJ

64 .0 kJ

80 .0 kJ

92 .0 kJ

$2 N H_{3} \to N_{2} + 3 H_{2}$
2 moles
The enthalpy of formation of ammonia
= - 46 kJ $m o l^{- 1}$
$Δ$ H of given reaction = -2 $\times$ (- 46) = 92 kJ

Q. The enthalpy of formation of ammonia is −462kJmol−1. The enthalpy change for the reaction 2NH3​−−−−−>N2​+3H2​ is