For the reactionH2F2g - H2g - F2g- U - -59-6 kJ mol-1 at 27- C-The enthalpy change for the above reaction is- kJ mol-1 -nearest integer-Given - R - 8-314 J K-1 mol-1-

H_2F_2(g) → H_2(g) + F_2(g) Δ U = –59.6 kJ mol^ 1 at 27^∘ C Δ H=Δ U+Δ n_gRT = 59.6+ 1×8.314×3001000 = 57.10 kJ mol^ 1 ...

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Difference between Internal Energy and Enthalpy

For the react...

Question

For the reaction
$H_{2} F_{2} (g) \to H_{2} (g) + F_{2} (g)$
$Δ U = - 59.6 k J m o l^{- 1}$ at $27^{\circ} C$ .
The enthalpy change for the above reaction is
(–) $k J m o l^{- 1}$ [nearest integer]

Given : $R = 8.314 J K^{- 1} m o l^{- 1}$ .

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Solution

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k = (6.5 \times 10^{12} s^{- 1}) e^{- 26000 K / T}

k J m o l^{- 1}

R = 8.314 J K^{- 1} m o l^{- 1}

298

- 42.0

m o l^{- 1}

R = 8.314

K^{- 1}

m o l^{- 1}

4 F e (s) + 3 O_{2} (g) \to 2 F e_{2} O_{3} (s)

- 550 J K^{- 1}

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- 165 k J m o l^{- 1}]

K

C (s) + O_{2} (g) - \to C O_{2} (g) Δ H^{\circ} = - 393.5 {kJ mol}^{- 1}

C (s) + \frac{1}{2} O_{2} (g) - \to C O (g) Δ H^{\circ} = - 110.5 {kJ mol}^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) - \to H_{2} O (g) Δ H^{\circ} = - 241.8 {kJ mol}^{- 1}

({in kJ mol}^{- 1})

C O_{2} (g) + H_{2} (g) ⟶ C O (g) + H_{2} O (g)

300 K

Δ_{c} H^{⊖} = - 601.70 k J m o l^{- 1}

R = 8.3 J K^{- 1} {mol}^{- 1}

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