Calculate equilibrium constant at 25C And 45C

Question

I’m given the equilibrium concentration of of chemical reaction at 25C [SO2]= 0.1 M, [O2] = 0.3 M and [SO3] = 0.5 M

2 SO2 (g) + O2 (g) <——> 2 SO3 (g).

I’m asking to find the equilibrium constant at 25C and also 45C?

I’m also the direction the equilibrium will shift at 49C?

Delta f H SO2= -296.81 kJ/mol ; Delta f H SOs= -395.77 kJ/mol

Answer 1

2SO₂(g) + O₂(g) <==> 2SO₃(g)

0.1.............0.3.................0.5.............Equilibrium concentrations

Write the equilibrium expression:

Keq = [SO₃]² / [SO₂]²[O₂]

Plug in values, and solve for Keq:

Keq = (0.5)² / (0.1)²(0.3)

Keq = 83.3

To find Keq at a different temperature, we'd need to us the van't Hoff equation, and for that we need to know ∆Hº. So we'll first find this value:

∆Hº = ∑∆HfProducts - ∑∆HfReactants

∆Hº= (2 x -395.77) - (2 x -296.81) = (-791.54) - (-593.62)

∆Hº = -197.92 kJ/mol

Van't Hoff equation:

ln(K2/K1) = -∆H/R (1/T2 - 1/T1)

K1 = 83.3

K2 = ?

∆H = -197.92 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol

T1 = 25ºC + 273 = 298K

T2 = 45ºC + 273 = 318K

ln (K2/83.3) = - (-197.92/0.008314) (1/318 - 1/298)

ln (K2/83.3) = 23806 x -1.2x10^-4

ln (K2/83.3) = -2.86

K2/83.3 = 0.0573

K2 = 4.77 = Keq @ 45ºC

NOTE: ∆Hº is negative, meaning the reaction is exothermic, so heat will appear on the product side of the reactxion. According to LeChatelier, the equilibrium will shift to the LEFT at 45º