Question

1) What is the change in entropy (ΔS) for the following reaction at 25 °C?

CS2(l) + 3O2(g) → CO2(g) + 2SO2(g) ΔS = ?

ΔS[CS₂] = 151 J / mol · K,  ΔS[O₂] = 205 J / mol · K, ΔS[CO₂] = 214 J / mol · K, and  ΔS[SO₂] = 248 J / mol · K

a.56.0 J / K

b.-56.0 J / K

c.111 J / K

d.111 J / K

2) What is the Gibbs Free Energy Value (ΔG) for the following reaction at 25 °C?

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)   ΔG = ?

ΔG[C₂H₄] = 68 kJ/mol,  ΔG[O₂] = 0.0 kJ/mol, ΔG[CO₂] = -394 kJ/mol, and  ΔG[H₂O] = -229 kJ/mol

a.657 kJ

b.-657 kJ

c.-1314 kJ

d.1314 kJ

3) What is the equilibrium constant, Keq, for the following reaction at 25 °C? (R = 8.314 J / mol · K)

2KClO₃(s) <=> 2KCl(s) + 3O₂(g) K = ?

ΔG[KClO₃] = -290 kJ/mol, ΔG[KCl] = -409 kJ/mol, and ΔG[O₂] = 0.0 kJ/mol

a.	5.23 x 1041
	b.	3.76 x 10-33
	c.	6.11 x 1016
	d.	3.03 x 10-8

4) Given the following thermodynamic data, calculate the Free Energy (ΔG) and then determine if the reaction is spontaneous or non-spontaneous at 25 °C. [C(Graphite) + 2H₂(g) → CH₄(g)   ΔG = ?]

ΔG[C(Graphite) = 0.0 kJ/mol, ΔG[H₂] = 0.0 kJ/mol, and ΔG[CH₄] = -51 kJ/mol

a.	ΔG = 25.5 kJ; Non-Spontaneous
	b.	ΔG = -25.5 kJ; Spontaneous
	c.	ΔG = 51 kJ; Non-Spontaneous
	d.	ΔG = -51 kJ: Spontaneous

Answer 1

1)

ΔS for a reaction is given as:

ΔS(reaction) = Σ(nΔS(products)) – ΣnΔS(reactants))

Here, n = number of moles / the stoichiometric coefficient from the balanced chemical reaction

And Σ sign denotes summation.

So, for the reaction : CS₂(l) + 3 O₂(g) = CO₂ (g) + 2 SO₂ (g)

ΔS = [{(2moles) ΔS(SO₂)} + {(1moles) ΔS(CO2)}] –[ {(1moles) ΔS (CS2)} + {(3moles) ΔS (O2)} ]

Putting the values:

ΔS = [{(2 mol)(248 J/mol.K) + (1 mol)(214 J/mol.K)] –[ (1 mol)(151 J/mol.K) + (3 mol)(205 J/mol.K) ]

Or, ΔS = -56.0 J/K

2)

The formula to calculate ΔG for the reaction is exactly the same as ΔS above, just replace ΔS by ΔG values.

So, formula is :

ΔG(reaction) = Σ(nΔG(products)) – ΣnΔG(reactants))

The reaction is : C₂H₄ (g) + 3 O₂ (g) = 2 CO₂ (g) + 2 H₂O (g)

ΔG= [{(2moles) ΔG(H2O)} + {(2moles) ΔG(CO2)}] –[ {(1moles) ΔG (C₂H₄)} + {(3moles) ΔG (O2)} ]

Or ΔG = [{(2 mol)(-229 kJ/mol) + (2 mol)(-394 kJ/mol)] –[ (1 mol)(68 kJ/mol) + (3 mol)(0.0 kJ/mol) ]

Or, ΔG = -1314 kJ

3)

Here, first calculate ΔG as done before. The find K from ΔG using : ΔG = -RTlnK ( T = temperature in K)

ΔG(reaction) = Σ(nΔG(products)) – ΣnΔG(reactants))

So, ΔG = [{(2 mol)(-409 kJ/mol) + (3 mol)(0.0 kJ/mol)] –[ (2 mol)(-290 kJ/mol) ]

Or, ΔG = -238 kJ

ΔG = -RTlnK. Here, R = 8.314 J/mol-K

So, the units in R are – J for ΔG and K for temperature.

So, convert ΔG to J : ( -238kJ) = - ( 238 x 10³ ) J [ Since 1 kJ = 10³ J]

Convert T to K : 25 C = ( 25 + 273 ) K = 298 K

Putting the values:

- (238 x 10³ ) = - (8.314)(298) ln K

Solving, K = 5.23 x 10⁴¹

4)

ΔG(reaction) = Σ(nΔG(products)) – ΣnΔG(reactants))

ΔG = [{(1 mol)(-51 kJ/mol)] –[ (2 mol)(-0.0 kJ/mol) + (1 mol)(-0.0 kJ/mol) ]

Or, ΔG = -51 kJ

If ΔG is negative, the reaction is spontaneous. If ΔG is positive, the reaction is not spontaneous.

So, this reaction is spontaneous.