Question

For the reaction

H₂(g) + Cl₂(g) 2 HCl(g)

G° = -189.8 kJ and S° = 20.0 J/K at 262 K and 1 atm.

This reaction is (reactant, product) favored under standard conditions at 262 K.

The standard enthalpy change for the reaction of 2.46 moles of H₂(g) at this temperature would be kJ.

Answer 1

For the reaction

H₂(g) + Cl₂(g) ==> 2 HCl(g)

G° = -189.8 kJ and S° = 20.0 J/K at 262 K and 1 atm.

•The reaction is product favoured under the standard condition at 262 K because of G° < 0.

•The standard enthalpy change for the reaction of 1 mole of H₂(g) at this temperature.

G° = H° -TS° (Gibbs equation)

-189.8 kJ = H° -(262 K)(20.0 J/K)

-189800 J = H° -5240 J

H° = -189800+5240 J = -184,560 J = -184.56 kJ/mol

The standard enthalpy change for the reaction of 1 mole of H₂(g) at this temperature is 184.56 kJ/mol.

So,

For 2.46 mol of H2(g):-

H° = (2.46 mol)(184.56 kJ/mol) = 454.02 kJ

(The standard enthalpy change for the reaction of 2.46 moles of H₂(g) at this temperature is 454.02 kJ.)