Find ΔrG for the following (in kJ mol-1) N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g)
The conditions for this reaction are:
Temp: 298k
P - NH3 = 0.95 bar
P - H2 = 1.95 bar
P - N2 = 1.25 bar
NH3(g) ?H ∙(kJ mol-1) = -45.9 ?G ∙(kJ mol-1) = -16.4 S ∙(J K-1 mol-1)192.8
N2(g) ?H ∙(kJ mol-1) = 0 ?G ∙(kJ mol-1) = 0 S ∙(J K-1 mol-1)191.6
H2(g) ?H ∙(kJ mol-1) = 0 ?G ∙(kJ mol-1) = 0 S ∙(J K-1 mol-1)130.7
=======================HINT============
You must find Q and ∆rG° then use
∆rG = ∆rG° + RT ln Q
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