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Enthalpy

For N2g + 3 H...

Question

For $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ , $Δ H$ is equal to:

A

Δ U + 2 R T

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B

Δ U - 2 R T

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C

Δ U - R T

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D

Δ U + R T

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Solution

The correct option is B $Δ U - 2 R T$

We know, $Δ H = Δ U + Δ n_{g} R T$
where $Δ H$ = change in enthalpy of the reaction
$Δ U$ = Change in internal energy
$Δ n_{g}$ = difference in moles of products and reactants = $2 - 4 = - 2$
$Δ H$ = $Δ U - 2 R T$

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Similar questions

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g); Δ H = - 22 k c a l

Activation energy,

E_{a}

for the given reaction is

70 k c a l

. Find the activation energy for

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

k c a l

.

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

For this reaction relate

Δ H

Δ U

.

Δ C_{p}

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

Q. The value of

Δ H - Δ E

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

:

Q. Assertion :For a reaction:

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

Reason: Enthalpy change is always greater than internal energy change.

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