The enthalpy change $(Δ H)$ for the reaction, $N_{2} (g) + {3 H}_{2} (g) ⟶ {2 N H}_{3} (g)$ is $- 92.38 k J$ at $298 K .$ The internal energy change $Δ U$ at $298 K$ is:

- 92.38 k J

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- 87.42 k J

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- 97.34 k J

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- 89.9 k J

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Solution

The correct option is A $- 92.38 k J$
$△ H = - 92.38 K J$
$△ T = 298 K - 298 K$
$= 0$
$△ U = △ H - R △ T$
$= - 92.38 - R (0)$
$= - 92.38$

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Similar questions

(Δ H)

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

- 92.38 k J

298 K

Δ U

298 K

(△ H)

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

△ U

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

Δ H > Δ U

2 {N H}_{3} (g) ⟶ N_{2} (g) + 3 H_{2} (g); Δ H > Δ E

Δ H > Δ U

Δ H =

Δ U =

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