Question

Version A 12. If AH = 75 kJ/mol and AS = 75J/mol K for an equilibrium reaction at 1400 K, describe the equilibrium constant K. (A) K=0 (B) 0<K<1 (C) K=1 (D) K>1 13. A perturbation of an equilibrium is observed to have caused additional products to form. The reaction is exothermic and has An = 2. Which of the following stresses could have caused the observation? (A) Increased Pressure (C) Removal of Product (E) None of these (B) Decreased temperature (D) Two of these 14. Kp = 12.0 for the following equilibrium at 298 K. A(g) + B(g) = AB(g) What is the value of K for this reaction at 298 K? (A) 0.00484 (B) 0.490 (C) 294 (D) 2.97 X 104

Answer 1

we know the equaion

delta G=delta H -T delta S

H=75 kj/mol=75000 j/mol

S=75 j/mol

T=1400 k

delta G=75000-1400 X75

=-30000 j/mol

we also know the equation

delta G0=delta G +RT lnK

as at equibilrium delta G0=0

so

delta G=-RT lnK

-30000=-RT lnK

ln K=30000/8.314 X 1400

ln K=2.577

as we can clearly say that K>1

13

this problem is based on le chaterial principle

when we inrease the pressure equilbilrium will shift to that side which contain less no of moles of gaseous species.his nullify effect of increase in pressure

as delta n=2

so moles of reactant is less than moles of product

hence reacion shift towards reactant side and les product is obtaine

so this is wrong option

since it is exothermic reaction and according to le chaterial exothermic reation is favoured by decreaing the temprature ..so this option is correct

if we decreasing the product the eaction shift towards product side according to le chaterial..hence this option is also correct

since two options are correct

so answer is option D

14.

we know the equation

Kp=Kc (RT)delta n of gaseous products

Kp=12

T=298 k

R=8.314

delta n=-1 (moles of product-moles of reactant)

12=KC x( 8.314 x 298)-1

Kc=2.97 x 104