Equilibrium constant

Question

Consider the reactions 5a and 5b, below, taking place at 25 °C, and answer the following questions.

5a H​2S​ O​3(​ g) ⟶ SO​2(​ g) + H​2​O(g) △Go​ ​ = +10 kJ mol-​ 1

5b H​2(​ g) + Cl​2(​ g) ⟶ 2 HCl(g) △Go​ ​ = -400 kJ mol-​1

(a)For each reaction, 5a and 5b, calculate the equilibrium constant, ​K​, and state if the equilibrium position lies to the right or the left. Recall that:

∆​G​° = –​RT​ ln ​K​ where ​R​ = 8.31 J mol-​ 1​ K-​ 1​ and ​T​ / K = 273 + ​T​ / °C

(b) which direction will each reaction shift upon an increase of pressure?

C)) Calculate the reaction quotient for reaction 5a under the following conditions: 25C(degree celcius); [SO​2]​ =0.01M;[H​2O​ ]=0.05M;[H​2​SO​3]​ =0.01M. In which direction will the reaction proceed:right or left?

Answer 1

H​2S​O​3( g) ⟶ SO​2(​ g) + H​2​O(g)

5a) △Go​ ​ = +10 kJ mol-​ 1

∆​G​° = –​RT​ ln ​K so plug in the numbers and solve for K

10 kJ/mole = -0.00831 kJ/mol-K x 298K x ln K (note to change R to kJ or change ∆G to joules)

Solve for K

If K >1, equilibrium will lie to the right and vice versa

Increasing pressure will shift reaction to the reactants because there are more moles of gas on product side

5b) H​2(​g) + Cl​2(​g) ⟶ 2 HCl(g) △Go​ ​ = -400 kJ mol-​1

Solve for K same as in (5a) above.

Increasing pressure will have no effect because number of moles of gas is same on both sides of equation.

C. H​2S​O​3( g) ⟶ SO​2(​ g) + H​2​O(g)

Q = [SO2][H2O]/[H2SO3] = (0.01)(0.05)/(0.01) = 20

Compare this to calculated K above and if it is >K reaction will proceed toward the reactants, and if it is <K, reaction will proceed toward products.