Question

Consider the following reaction occurring in a 1L container:   2A --> A₂

The enthalpy for this reaction is a function of temperature and is given as Delta H(T) = -10T^-1/6 KJmol^-1K^-1

a) Use the Van't Hoff equation to calculate the value of the equilibrium constant, K, at 425 degrees C if K = 0.5 at 25 degrees C.

b) Beginning with 2 mol of A, calculate [A₂] at 25 degrees C.

c) Beginning with 2 mol of A, calculate [A₂] at 425 degrees C.

Answer 1

According to van't Hoff equation:

ln(K₂/K₁) = -∆H/R[1/T₂ -1/T₂]

K₂ = equilibrium constant at T₂ temperature

K₁ = equilibrium constant at T₁ temperature

R = gas constant = 8.314J/mol.K

425 oC = 425 + 273 = 698 K

∆H(T) = (-10T^-1/6)KJmol^-1K^-1 given in question

= (-10 x 698^{- 1/6}) KJmol^-1K^-1 = -3.35756 kJ/mol = -3357.56 J/mol

T₁ = 25 ^oC = 298 K

ln(K₂/K₁) = -3357.56 J/mol /8.314 J/mol.K[1/698 K - 1/298K] =

(K₂/K₁) = 2.174

K₂ = 0.5 x 2.174 = 1.087

b)

2A <--> A₂

2 mol 0 initially

2-2x x equilibrium

K = [A₂]/[A]²

0.5 = x / [2-2x]²

2x² - 5x +2 = 0

this quadratic equation will give two roots;

x₁ = 2

x₂ = 0.5

x₂ is accepted, X₁ = 2 will make reactant negative (2-2x) = -2

A remaining = 2 - 2x = 2-0.5 x 2 = 1 mole

B = x = 0.5 mole

c)

at 425 oC , K = 1.087

1.087 = x / [2-2x]²

4.348x² - 9.696x + 4.348 = 0

x = 1.6 , 0.62

0.62 is accepted

A = 2-2x = 2-2 x 0.62 = 0.76

B = x = 0.62