If the enthalpy change for the reaction is 2NH3(g)arrow N2(g)+3H2(g) is 92 kJ mol-1 the enthalpy of formation of ammonia is:

Question

If the enthalpy change for the reaction is 2NH3(g)arrow N2(g)+3H2(g) is 92 kJ mol-1 the enthalpy of formation of ammonia is: (A)  -23.0 kJ mol-1  (B)  -46 kJ mol-1  (C)  -92.0 kJ mol-1  (D)  46 kJ mol-1

Tardigrade Admin · Accepted Answer

2 NH 3(g) arrow N 2(g)+3 H 2(g) ; Δ H =+92 kJ N 2(g)+3 H 2(g) arrow 2 NH 3(g) ; Δ H =-92 kJ (1/2) N 2(g)+(3/2) H 2(g) arrow NH 3(g) ; Δ H =-46 kJ Hence, enthalpy of formation of NH 3 is -46 kJ mol -1

Q. If the enthalpy change for the reaction is $2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$ is $92 k J m o l^{- 1}$ the enthalpy of formation of ammonia is:

$- 23.0 k J m o l^{- 1}$

$- 46 k J m o l^{- 1}$

$- 92.0 k J m o l^{- 1}$

$46 k J m o l^{- 1}$

$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g); Δ H = + 92 k J$
$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g); Δ H = - 92 k J$
$\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to N H_{3} (g); Δ H = - 46 k J$
Hence, enthalpy of formation of $N H_{3}$ is $- 46 k J m o l^{- 1}$

Q. If the enthalpy change for the reaction is 2NH3​(g)→N2​(g)+3H2​(g) is 92kJmol−1 the enthalpy of formation of ammonia is:

−23.0kJmol−1

−46kJmol−1

−92.0kJmol−1

46kJmol−1