The heat of formation of NH3(.g.) is -46 kJmol- 1 . The Δ H (in kJ) of the reaction is 2NH3(.g.) arrow N2(.g.)+3H2(.g.)

Question

The heat of formation of NH3(.g.) is -46 kJmol- 1 . The Δ H (in kJ) of the reaction is 2NH3(.g.) arrow N2(.g.)+3H2(.g.) (A) 46 (B) - 46 (C) 92 (D) - 92

Tardigrade Admin · Accepted Answer

(1/2)N2(.g.)+(3/2)H2(.g.) arrow NH3(.g.) Δ Hr=-46kJ/mol 2NH3(.g.) arrow N2(.g.)+3H2(.g.) Δ Hr=-2Δ Hr =-2(.-46.)=92kJ

Q. The heat of formation of $N H_{3} (g)$ is $- 46 k J m o l^{- 1}$ . The $Δ H$ (in kJ) of the reaction is

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

46

$-$ 46

92

$-$ 92

$\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to N H_{3} (g)$

$Δ H_{r} = - 46 k J / m o l$

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

$Δ H_{r} = - 2Δ H_{r}$

$= - 2 (- 46) = 92 k J$

Q. The heat of formation of NH3​(g) is −46kJmol−1 . The ΔH (in kJ) of the reaction is 2NH3​(g)→N2​(g)+3H2​(g)

46

− 46

92

− 92

21​N2​(g)+23​H2​(g)→NH3​(g) ΔHr​=−46kJ/mol 2NH3​(g)→N2​(g)+3H2​(g) ΔHr​=−2ΔHr​ =−2(−46)=92kJ

Q. The heat of formation of $N H_{3} (g)$ is $- 46 k J m o l^{- 1}$ . The $Δ H$ (in kJ) of the reaction is

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

$-$ 46

$-$ 92

$\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to N H_{3} (g)$

$Δ H_{r} = - 46 k J / m o l$

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

$Δ H_{r} = - 2Δ H_{r}$

$= - 2 (- 46) = 92 k J$