Heat of formation of ${N H}_{3}$ at $4298 K$ and constant pressure is $- 11 k c a l {m o l}^{- 1}$ . Calculate heat of formation of ${N H}_{3}$ at $398 K$ and constant pressure. What are corresponding values at constant volume?

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Solution

$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$
Here $△ n_{g} = d i f f e r e n c e b e t w e e n g a s e o u s m o l e s o f p r o d u c t s a n d r e a c t a n t s$
$= 2 - (3 + 1) = - 2$
$∵ △ V = △ H - △ n_{g} R T$
At $T = 4298 K, △ V = (- 11) - (- 2) \times 2 \times 4298 \times 10^{- 3} k c a l m o l^{- 1} = 6.192 k c a l m o l^{- 1}$ =heat of formation at constant volume( $T = 4298 K$ ).
$△ H_{2} - △ H_{1} = C_{p} △ T$
$C_{p} = 4 R$ for polyatomic gas $N H_{3}$
$∴ (- 11) - △ H_{1} = 4 \times 2 \times (4298 - 398) \times 10^{- 3}$
$△ H_{1} = - 11 - 31.2 = - 42.2 k c a l m o l^{- 1}$
At $T = 398 K V = (- 42.2) - (- 2) \times 2 \times 398 \times 10^{- 3} = - 40.608 k c a l m o l^{- 1}$

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Heat of formation of NH3 at 4298K and constant pressure is −11kcalmol−1. Calculate heat of formation of NH3 at 398K and constant pressure. What are corresponding values at constant volume?

Heat of formation of ${N H}_{3}$ at $4298 K$ and constant pressure is $- 11 k c a l {m o l}^{- 1}$ . Calculate heat of formation of ${N H}_{3}$ at $398 K$ and constant pressure. What are corresponding values at constant volume?