The heat of reaction $(Δ H)$ for the formation of ${N H}_{3}$ according to the reaction $N_{2} + {3 H}_{2} ⇌ {2 N H}_{3}$ at $27^{o} C$ was found to be $- 91.94 k J$ . The heat of reaction $Δ H$ at $50^{o} C$ is 80KJ.
$[R = 8.31 J / m o l / K]$

True

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False

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Solution

The correct option is B False
$N_{2} 3 H_{2} ⇌ 2 {N H}_{3}$
$Δ H_{2} = - 91.94 k J T = 300 K$
$(Δ H_{2})_{T_{2}} - (Δ H_{2})_{T_{1}} : Δ C_{p_{r}} (T_{2} - T_{1})$
$C_{p} = \frac{T R}{2} + R, g :$ degree of freedom
$= 3 :$ Mono atomic gas
$= 5 :$ Distance & poly atomic lines gas
$= 6 :$ Poly atomic non line
$Δ C_{p_{r}} = \sum (C_{p_{p r o d u c t}} \times s t o s c o e f f i c i e n t) - \sum (C_{p_{r e a c t a n t}} \times s t i o c o e f f i c i e n t)$
$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$
$Δ C_{p_{r}} = 2 (4 R) - 3 (\frac{7 R}{2}) - (\frac{5 R}{2})$
$= - 5 R$
$(Δ H_{2})_{T_{2}} = (Δ H_{2})_{T_{1}} - 5 R (23)$
$= - 91.94 k J - 0.956 k J$
$= - 92.89 k J$

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Similar questions

N_{2} + 3 H_{2} \to 2 {N H}_{3}

27^{o} C

- 91.94 k J

50^{o} C

P

27^{o} C

N_{2}, H_{2}

{N H}_{3}

28.45, 28.32

37.07 k J

N_{2} + 3 H_{2} ⟶ 2 N H_{3}

(Δ H - Δ U)

N_{2} + 3 H_{2} ⇌ 2 N H_{3}; Δ H =

N_{2} + 3 H_{2} ⇌ 2 {N H}_{3}; Δ H = - v e

T i C L_{4} (g) + 2 H_{2} O (g) \to T i O_{2} (g) + 4 H C l (g)

Δ H_{f}^{\circ} T i C L_{4} (g) = - 763.2 k J / m o l

Δ H_{f}^{\circ} T i O_{2} (g) = - 944.7 k J / m o l

Δ H_{f}^{\circ} H_{2} O (g) = - 241.8 k J / m o l

Δ H_{f}^{\circ} H C l (g) = - 92.3 k J / m o l

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