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

`\Delta G^(0)` = -457.9 kJ and reaction is product favored.

Step-by-step 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* 10^(3))J-(273K* -365.6J/K)=-583291.2J`

So, for the reaction of 1.57 moles of
`NH_(3)`,
`\Delta G^(0)=((1.57)/(2))* -583291.2J=-457883.592J=-457.9kJ`

As,
`\Delta G^(0)` is negative therefore reaction is product favored under standard condition.