In a closed flask at 600 K, one mole of X_2Y_4(g) attains equilibrium as given below : X _2 Y _4(g) 2 XY _2(g) At equilibrium, 75\% X_2Y_4(g) was dissociated and the total pressure is 1 atm. The magnitude of _r G^ (in kJ mol^ -1 ) at this temperature is __________. (Nearest Integer) (Given : R = 8.3 J mol^ -1 K^ -1 ; 10 = 2.3, 2 = 0.3, 3 = 0.48, 5 = 0.69, 7 = 0.84)
A. A
The given equilibrium reaction is: X _2 Y _4(g) 2 XY _2(g) Initial moles: 1 Degree of dissociation, = 0.75 Moles of X _2 Y _4 at equilibrium = 1 - = 1 - 0.75 = 0.25 Moles of XY _2 at equilibrium = 2 = 2 0.75 = 1.5 Total moles at equilibrium = 0.25 + 1.5 = 1.75 Given total pressure, P = 1 atm. Partial pressure of X _2 Y _4, P_ X _2 Y _4 = 0.25 1.75 1 = 1 7 atm Partial pressure of XY _2, P_ XY _2 = 1.5 1.75 1 = 6 7 atm The equilibrium constant K_p is given by: K_p = (P_ XY _2 )^2 P_ X _2 Y _4 = ( 6 7 )^2 1 7 = 36 7 The standard Gibbs free energy change is: _r G^ = -RT K_p = -2.3 RT K_p Calculating K_p: ( 36 7 ) = 36 - 7 = (2^2 3^2) - 7 ( 36 7 ) = 2 2 + 2 3 - 7 ( 36 7 ) = 2(0.3) + 2(0.48) - 0.84 = 0.6 + 0.96 - 0.84 = 0.72 Substituting the values into the _r G^ equation: _r G^ = -2.3 8.3 600 0.72 _r G^ = -8246.88 J mol ^ -1 = -8.24688 kJ mol ^ -1 The magnitude of _r G^ is 8.24688 kJ mol ^ -1 . Rounding to the nearest integer, we get 8. Answer: 8
Related: Chemistry — Chemical Equilibrium · All PYQ Banks