(NH3) = 0.0312 mols/5 L = approx 0.006
(CO2) = 0.0156 mols/5 L = approx 0.003
Note that concn NH4CO2NH2 doesn't matter since it doesn't appear in the Q expression.
NH4CO2NH2(s) ⟶ 2 NH3(g) + CO2(g)
dG = -RTlnQ
dG = -RTln(NH3)^2*(CO2)
Substitute and solve for Q and dG.
BTW, what does r stand for in delta r G?
Note that my calculations are approx. You need to recalculate all.
Also note that dG will be in joules and you will need to convert to kJ.
For the reaction below, the thermodynamic equilibrium constant is K = 2.30×10−4 at 25 °C.
NH4CO2NH2(s) ⟶ 2 NH3(g) + CO2(g)
Suppose that 0.0156 moles of NH4CO2NH2, 0.0312 moles of NH3, and 0.0156 moles of CO2 are added to a 5.00 L container at 25 °C.
(a) What are Q and ΔrG (kJ mol−1) for the initial reaction mixture?
2 answers
r stands for reaction! thanks for ur help
5 answers