You visited us 1 times! Enjoying our articles? Unlock Full Access!

Hess' Law

At 1000 K, ...

Question

At $1000 K$ , from the data :
$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ ; $△ H = - 123.77 k J m o l^{- 1}$

Substance $N_{2}$ $H_{2}$ $N H_{3}$
$P / R$ $3.5$ $3.5$ $4$

Calculate the heat of formation of $N H_{3}$ at $300 K$ .

- 44.42 k J m o l^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 88.85 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 44.42 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 88.85 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $- 44.42 k J m o l^{- 1}$
Fro, Kirchhoff's equation
$△ H_{2} (1000 K) = △ H_{1} (300 K) + △ C_{p} (1000 - 300)$
Here, $△ H_{2} (1000 K) = - 123.77 k J m o l^{- 1}$
$△ H_{1} (300 K) = ?$
$△ C_{p} = 2 C_{p} (N H_{3}) - [C_{p} (N_{2}) + 3 C_{p} (H_{2})]$
$= - 6 R$
$= - 6 \times 8.314 \times 10^{- 3} k J$
$∴ - 123.77 = △ H_{1} (300 K) + 6 \times 8.314 \times 10^{- 3} \times 700$
or $△ H_{1} (300 K) = - 88.85 k J$
For two moles of $N H_{3}$
$∴ △ H_{f} (N H_{3}) = \frac{△ H_{1} (300 K)}{2}$
$= - \frac{88.85}{2}$
$= - 44.42 k J m o l^{- 1}$

Suggest Corrections

Similar questions

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

Δ H^{\circ} = - 123.77 K J m o l^{- 1}

727^{\circ} C

C_{p}

K J m o l^{- 1}

27^{\circ} C

N_{2}

H_{2}

N H_{3}

C_{p}

The equilibrium constant $K_{p}$ for the reaction $N_{2}$ (g) + 3 $H_{2}$ (g) $⇋$ $2 N H_{3}$ (g) is:

2

N_{2}

3

H_{2}

3 L

N H_{3}

1 / 3 m o l / L

K_{c}

N H_{3}

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

The equilibrium constant $K_{c}$ for the reaction $N_{2}$ (g) + 3 $H_{2}$ (g) $⇋$ 2 $N H_{3}$ (g) is:

K_{p}

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

1.64 \times 10^{- 4}

400^{o} C

0.144 \times 10^{- 4}

500^{o} C

1

{N H}_{3}

Join BYJU'S Learning Program