Question

For the reaction below, determine the equilibrium partial pressure (in atm) of HBr at 252.70 K if the initial pressure of H₂ is 2.83 atm. Assume that ΔH and ΔS do not vary with temperature. Report your answer to three significant figures.



LiBr (s) + H₂ (g) ⇌ LiH (s) + HBr (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

for the question first we need to find the values of r and r

i.e enthalpy of the reaction and the entropy of the reaction respectively.

r = (Sum of the enthalpies of the products) - (sum of the enthalpies of the reactants)

= {(-90.50)+(-36.29)} - {(-351.20)}

=224.41KJ/mol

now to calculate r we follow the same procedure,

r = (sum of the entropies of products) - (sum of entropies of the reactants)

={(20+198.70)} - {(74.30+130.68)}

=13.72

now we calculate the value of

i.e Gibbs free energy

= (224.41 X 1000) - (252.7X13.72)

=220942.956

using this we calculate the value of K

therefore,

therefore we konw that K=Kp

hence Kp= {partial presssure of (HBr)} / {partial pressure of (H2)}

therefore partial pressure of HBr atm