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What is the standard Gibbs energy change for the reaction at 298 K, if standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are 382.64 kJ $mol^{-1}$ and -145.6 $JK^{-1}mol^{-1}$, respectively.


(a) -523.2 kJ $mol^{-1}$
(b) -221.1 kJ $mol^{-1}$
(c) -339.3 kJ $mol^{-1}$
(d) -439.3 kJ $mol^{-1}$
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Answer: -339.3 kJ $mol^{-1}$
$NH_3 (g) + \frac{3}{2}O_2(g) \rightarrow \frac{1}{2}N_2(g) + \frac{3}{2}H_2O(l)$
$T=298 K$
$\Delta H = -382.64 kJ mol^{-1}$
$\Delta S = -145.6 JK^{-1}mol^{-1} = -0.1456 kJ K^{-1}mol^{-1}$
$\therefore \Delta G = \Delta H - T \Delta S = -382.64 kJmol^{-1} - (298 K) \times (0.1456 kJK^{-1}mol^{-1})$
$\Delta G = -382.64 kJmol^{-1} + 43.3888 kJ mol^{-1}= - 339.25 kJmol^{-1}$
answered Feb 25, 2014 by mosymeow_1
 

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