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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Calculate the standard free energy change, $ \Delta G^o$ for the reaction \[\] $H_2(g) + I_2(g) \rightarrow 2HI(g)$ \[\] Given that the standard enthalpy change and the standard entropy change for the reaction are $52 \: kJ\: and\: 165.2 \: J/K$ respectively.

$\begin {array} {1 1} (a)\;2.7 kJ & \quad (b)\;-2.7 kJ \\ (c)\; 101.23 kJ & \quad (d)\;-109.23 kJ \end {array}$

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$ \Delta H^o = 52 \: kJ\: \: \: \: \: \: \: \: \Delta S^o = 165.2\: J/K \: \: \: \: \: \: \: \: T = 298\: K$
$\Delta G^o = \Delta H^o - T \Delta S^o = \large\frac{52 – 298 \times 165.2}{1000 }$$= 2.7\: kJ$
Ans : (a)
answered Mar 13, 2014 by thanvigandhi_1
 

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