Question

For the reaction below, determine the equilibrium partial pressure (in atm) of H₂ at 714.10 K if the initial pressure of HBr is 1.24 atm. Assume that ΔH and ΔS do not vary with temperature. Report your answer to three significant figures.



LiH (s) + HBr (g) ⇌ LiBr (s) + H₂ (g)  

	ΔHf° (kJ/mol)		S° (J mol-1 K-1)
LiBr	-351.20		74.30
H2	0.00		130.68
LiH	-90.50		20.00
HBr	-36.29		198.70

Answer 1

Equations to be used delta G degree = -RT l n k _p (1) delta G degree _reaction = delta G degree _r times n = sigma delta G degree _product (f) - sigma delta G degree _reactants (f0 (2) delta G _degree = delta H degree - T delta s degree (3) 1)delta G degree (f) (Li Br) = delta H degree _f (Li Br) - T delta s degree (Li Br) downarrow Gibb's Free energy of formation = -351.20 KJ/mol - 714.10 K (74.30 J/K mol) = -351.2 KJ/mol - 53057.63 J/mol times (10^-3 KJ/J) = -351.2 KJ/mol - 53.058 KJ/mol = -404.258 KJ/mol 2)delta G degree _f (H _2) = delta H degree _f 9H _2) - T delta s degree (H _2) = 0 - 714.10 K (130.68 J/K mol) = -93318.59 J/mol times 10^-3 KJ/J = -93.318 KJ/mol 3) delta degree _f (Li H) = delta degree _f (Li H) - T delta s degree (Li H) = -90.50 - 714.10 K (20.0 J/K mol) = -104.78 KJ/mol