Problem: Given the following information determine the reaction Gibbs energy for the formation of hydrazine (N2H4) at 298 K 2NH3(g) → N2H4(l) + H2(g) NH3 :    ΔH°f = -46.11 kJ mol-1           ΔS° = 192.45 J mol-1 K-1N2H4 :  ΔH°f = 50.63 kJ mol-1           ΔS° = 121.21 J mol-1 K-1H2 :       ΔH°f = 0                                ΔS° = 130.7 J mol-1 K-1Answer to TWO decimals places in kJ mol-1

We can use the following equation to solve for ΔG˚_rxn:

$\boxed{\mathbf{\Delta G}{\mathbf{°}}_{\mathbf{rxn}}\mathbf{=}\mathbf{\Delta H}{\mathbf{°}}_{\mathbf{rxn}}\mathbf{-}\mathbf{T\Delta S}{\mathbf{°}}_{\mathbf{rxn}}}$

For this problem, we need to do the following steps:

Step 1: Calculate ΔH˚_rxn.

Step 2: Calculate ΔS˚_rxn.

Step 3: Use ΔH˚_rxn and ΔS˚_rxn to calculate for ΔG˚_rxn.

Step 1: We can use the following equation to solve for ΔH˚_rxn:

$\boxed{\mathbf{\Delta H}{\mathbf{°}}_{\mathbf{rxn}}\mathbf{=}\mathbf{\Delta H}{\mathbf{°}}_{\mathbf{f}\mathbf{,}\mathbf{ }\mathbf{prod}}\mathbf{-}\mathbf{\Delta H}{\mathbf{°}}_{\mathbf{f}\mathbf{,}\mathbf{ }\mathbf{react}}}$

Note that we need to multiply each ΔH˚_f by the stoichiometric coefficient since ΔH˚_f is in kJ/mol. 

Also, note that ΔH˚_f for elements in their standard state is 0.