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

FREE Expert Solution

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

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

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:

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.

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Problem Details

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-1

N2H4 :  ΔH°f = 50.63 kJ mol-1           ΔS° = 121.21 J mol-1 K-1

H2 :       ΔH°f = 0                                ΔS° = 130.7 J mol-1 K-1

Answer to TWO decimals places in kJ mol-1