H 2 (g) + Cl 2 (g) -------> 2 HCl (g) Appendix 4: H2 (g): H o = 0, S o = 131 J/K
ID: 511499 • Letter: H
Question
H2 (g) + Cl2 (g) -------> 2 HCl (g)
Appendix 4: H2 (g): Ho = 0, So = 131 J/K * mol, Go = 0. Cl2 (g): Ho = 0, So = 223 J/K * mol, Go = 0. HCl (g): Ho = -92 Kj/mol, So = 187 J/K * mol, Go = -95 Kj/mol
a) Calculate Ho, So, Go and equilibrium constant K (at 298 K) using data in Appendix 4 of Zumdahl textbook
b) If H2 (g), Cl2 (g), and HCl (g) are placed in a flask such that the pressure of each gas is 1 atm, in which direction will the system will shift to reach equilibrium at 25oC?
Explain please.
Thanks!
Explanation / Answer
a)H(g) +cl2(g)-› 2HCL(g)
Given data
H2 (g): Ho=0, So = 131 J/K * mol, Go = 0. Cl2 (g): Ho = 0, So = 223 J/K * mol, Go = 0. HCl (g): Ho = -92 Kj/mol, So = 187 J/K * mol, Go = -95 Kj/mol
DeltaH= DeltaH(product)-Delta H(reactant)
=2*(-92)-0-0=184 KJ/mol
Delta S= Delta S(product)- Delta S(reactant)
=2*187-223-131 =20J/K mol
Delta G=Delta G(product)- Delta G(reactant)
=-95*2-0-0= -190KJ/mol
Equilibrium constant k
Delta G= -RT lnk
k= e^-(DeltaG/RT)
k= e^-(-190*10^3/8. 314*298)
k =2. 0192*10^33
b) when H2g, cl2g and HClg placed in a flask with 1atm pressure, the system will shift to forward direction and will occur mostly to the completion of reaction. It will use almost all the reactants or exhaust due to high value of k.
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