Question

For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the reaction of 1.57 moles of NH3(g) at 273 K, 1 atm would be kJ. This reaction is (reactant, product) favored under standard conditions at 273 K. Assume that H° and S° are independent of temperature.

Answer 1

Answer:

$\Delta G^{0}$ = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of $NH_{3}$

Standard free energy change of the reaction ($\Delta G^{0}$) is given as:

           $\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}$   , where T represents temperature in kelvin scale

So, $\Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J$

So, for the reaction of 1.57 moles of $NH_{3}$, $\Delta G^{0}=(\frac{1.57}{2})\times -583291.2J=-457883.592J=-457.9kJ$

As, $\Delta G^{0}$ is negative therefore reaction is product favored under standard condition.