Nitric oxide reacts with chlorine to form nocl. the data refer to 298 k. 2no(g) + cl2(g) → 2nocl(g) substance: no(g) cl2(g) nocl(g) ∆h°f (kj/mol): 90.29 0 51.71 ∆g°f (kj/mol): 86.60 0 66.07 s°(j/k·mol): 210.65 223.0 261.6 what is the value of ∆g° fo this reaction at 550 k?

- 10.555 kJ/mol.

Explanation:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).

∆H°rxn is the standard enthalpy change of the reaction (J/mol).

T is the temperature of the reaction (K).

∆S°rxn is the standard entorpy change of the reaction (J/mol.K).

∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants

∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) = (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.

∵  ∆S°rxn = ∑∆S°products - ∑∆S°reactants

∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) = (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) = - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

∴ ∆G°rxn = ∆H°rxn - T∆S°rxn = (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = - 10.555 kJ/mol.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2