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Standard XII

Chemistry

Enthalpy

If the standa...

Question

If the standard internal energy change for the reaction $O F_{2} (g) + H_{2} O (g) ⟶ O_{2} (g) + 2 H F (g)$ , at 298 K is $x$ $k J$ .
[Given standard enthalpies of formation in $k J m o l^{- 1}$ are $O F_{2} (g) = + 20; H_{2} O (g) = - 250 a n d H F (g) = - 270$ .]
Then $- 100 x$ is

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Solution

$Δ H_{r x n}^{o} = Δ H_{O_{2}}^{o} + 2 Δ H_{H F}^{o} - [Δ H_{O F_{2}}^{o} + Δ H_{H_{2} O}^{o}]$
$Δ H_{r x n}^{o} = 0 + 2 (- 270) - [+ 20 - 250] = - 310 k J = - 310000 J$
$Δ E_{r x n}^{o} = Δ H_{r x n}^{o} - Δ n R T$
$Δ E_{r x n}^{o} = - 310000 - (1 \times 8.314 \times 298) = - 312477 J = - 312.47 k J$
$- 100 x = 31247$

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Similar questions

Q. What will be the standard internal energy change for the reaction at 298 K?

O F_{2 (g)} + H_{2} O_{(g)} \to O_{2 (g)} + 2 H F_{(g)}; △ H^{0}

= - 310 kJ.

Q. What will be the standard enthalpy change for the reaction

O F_{2} (g) + H_{2} O (g) \to O_{2} (g) + 2 H F (g)

at 298K if enthalpies
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O F_{2} (g), H_{2} O (g)

and

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are +20, -250 and -270kJ

m o l^{- 1}

respectively.

Q. Calculate the standard enthalpy change and standard internal energy change for the following reaction at 300K :

O F_{2} (g) + H_{2} O (g) \to O_{2} (g) + 2 H F (g)

Given that the standard enthalpy of formation of

O F_{2}

H_{2} O

and

H F

are

23 k j m o l^{- 1}

- 241.8 k j m o l^{- 1}

and

- 268.6 k j m o l^{- 1}

respectively.

Q. Find the

Δ H

of the following reaction.

O F_{2} (g) + H_{2} O (g) ⟶ O_{2} (g) + 2 H F (g)

. Average bond energies of

O - F

O - H

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and

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44

111

118

and

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Standard enthalpy and standard entropy for the oxidation of $N H_{3}$ at $298$ K are $- 382.64$ $kJ {mol}^{- 1}$ and $- 145.6$ $J {mol}^{- 1} K^{- 1}$ respectively. Standard free energy change (in ${kJ mol}^{- 1}$ ) for the same reaction at $298$ K is:

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If the standard internal energy change for the reaction OF2(g)+H2O(g)⟶O2(g)+2HF(g), at 298 K is x kJ. [Given standard enthalpies of formation in kJmol−1 are OF2(g)=+20;H2O(g)=−250andHF(g)=−270.]Then −100x is

ΔHorxn=ΔHoO2+2ΔHoHF−[ΔHoOF2+ΔHoH2O]ΔHorxn=0+2(−270)−[+20−250]=−310kJ=−310000J ΔEorxn=ΔHorxn−ΔnRT ΔEorxn=−310000−(1×8.314×298)=−312477J=−312.47kJ −100x=31247

If the standard internal energy change for the reaction $O F_{2} (g) + H_{2} O (g) ⟶ O_{2} (g) + 2 H F (g)$ , at 298 K is $x$ $k J$ .
[Given standard enthalpies of formation in $k J m o l^{- 1}$ are $O F_{2} (g) = + 20; H_{2} O (g) = - 250 a n d H F (g) = - 270$ .]
Then $- 100 x$ is