(Solved): For the reaction of hydrazine, \( \mathrm{N}_{2} \mathrm{H}_{4} \), with hydrogen to produce ammoni ...

For the reaction of hydrazine, \( \mathrm{N}_{2} \mathrm{H}_{4} \), with hydrogen to produce ammonia \( \Delta G^{\circ}=-193 \mathrm{~kJ} \). \[ \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) \] Calculate \( \Delta G \) at \( 298 \mathrm{~K} \) when the reactants and product are at the following partial pressures: \( 0.01 \) \( \operatorname{atm} \mathrm{N}_{2} \mathrm{H}_{4}, 0.01 \mathrm{~atm} \mathrm{H}_{2} \), and \( 1.00 \times 10^{-3} \mathrm{~atm} \mathrm{NH}_{2} \). \( -125 \mathrm{~kJ} / \mathrm{mol} \) \( -169 \mathrm{~kJ} / \mathrm{mol} \) \( -204 \mathrm{~kJ} / \mathrm{mol} \) \( -233 \mathrm{~kJ} / \mathrm{mol} \)