The value of - H - U for the following reaction at 37- C will be2 NH 3 g - N 2 g -3 H 2 g A- 51-54 J mol -1B- -51-54 J mol -1C- 51-54 kJ mol -1D- 5-154 kJ mol -1

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Internal Energy

The value of ...

Question

The value of $Δ H - Δ U$ for the following reaction at $37^{o}$ C will be
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$

51.54 {kJ mol}^{- 1}

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51.54 {J mol}^{- 1}

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- 51.54 {J mol}^{- 1}

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5.154 {kJ mol}^{- 1}

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Solution

The correct option is D $5.154 {kJ mol}^{- 1}$
We know, $Δ H = Δ U + P Δ V$ = $Δ U + Δ n_{g} R T$
where $Δ H$ = Change in enthalpy
$Δ U$ = Change in internal energy
$Δ n_{g}$ = $n_{p r o d u c t s} - n_{r e a c t a n t s}$
$Δ n_{g}$ = $4 - 2$ = 2
So, $Δ H$ - $Δ U$ = 2RT = $2 \times 8.314 \times 310 = 5154.68 J m o l^{- 1} = 5.154 k J m o l^{- 1}$

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The value of $Δ H - Δ U$ for the following reaction at $27^{\circ}$ will be : $2 N H_{3 (g)} \to N_{2 (g)} + 3 H_{2 (g)}$

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

$(Δ H - Δ U)$ for the formation of carbon monoxide ( $C O$ ) from its elements at $298 K$ is ( $R = 8.314 J K^{- 1} m o l^{- 1}$ )

For the reaction $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), Δ H = - 285.8 k J m o l^{- 1} Δ S = - 0.163 k J m o l^{- 1} K^{- 1}$ . What is the value of free energy change $27^{\circ} C$ for the reaction.

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

For this reaction relate

Δ H

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Δ U

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Internal Energy

Standard XII Chemistry

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The value of ΔH−ΔU for the following reaction at 37o C will be 2NH3(g)→N2(g)+3H2(g)

The correct option is D 5.154 kJ mol−1We know, ΔH=ΔU+PΔV = ΔU+ΔngRT where ΔH = Change in enthalpy ΔU = Change in internal energy Δng = nproducts−nreactants Δng = 4−2 = 2 So, ΔH - ΔU = 2RT = 2×8.314×310=5154.68 Jmol−1=5.154 kJmol−1

The value of $Δ H - Δ U$ for the following reaction at $37^{o}$ C will be
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$