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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Calculate standard internal energy change($ \Delta E^{\circ}$) for the reaction \[\] $OF2(g) + H_2O(g) \rightarrow O_2(g) + 2HF(g)$ \[\] Given standard enthalpies of formation $ \Delta H_f^{\circ}\: of\: OF_2(g), H_2O(g)\: and\: HF(g)$ are23, -241.8 and -268.6 kJ/mol respectively.

$\begin {array} {1 1} (A)\;320.877\: kJ & \quad (B)\;-318.4 \: kJ \\ (C)\;-320.877 \: kJ & \quad (D)\;318.4\: kJ \end {array}$

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$ \Delta H^{\circ} = \sum \Delta H_f^{\circ} (products) - \sum \Delta H_f^{\circ} (reactants) $
$= 2(-268.6\: kJ) – {(23\: kJ) + (-241.8\: kJ)}$
$= -537.2 \: kJ + 218.8\: kJ = -318.4\: kJ$
$ \Delta n = 3-2 = 1\: mol$
$ \Delta H^{\circ} = \Delta E^{\circ} + (\Delta n)RT$
$ \Rightarrow \Delta E^{\circ} = \Delta H^{\circ} – (\Delta n)RT$
$= -318.4\: kJ – (1 \: mol)(8.314\: \Delta \: 10^{-3} \: kJ/Kmol)(298 \: K)$
$= -318.4\: kJ – 2476\: J = -318.4 \: kJ – 2.477\: kJ = -320.877\: kJ$
Ans : (C)
answered Mar 16, 2014 by thanvigandhi_1
 

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