Determine the entropy change for the combustion of gaseous propane, C3H8, under standard state conditions to give gaseous carbon dioxide and water.

ΔS°reaction = 100.9 J K⁻¹ (mol C₃H₈)⁻¹

Explanation:

The equation for the reaction is given as;

C₃H₈(g) + 5O₂(g) → 4H₂O(g) + 3CO₂(g)

In order to determine the entropy change, we have to use the entropy valuues for the species in the reaction. This is given as;

S°[C₃H₈(g)] = 269.9 J K⁻¹ mol⁻¹

S°[O₂(g)] = 205.1 J K⁻¹ mol⁻¹

S°[H₂O(g)] = 188.8 J K⁻¹ mol⁻¹

S°[CO₂(g)] = 213.7 J K⁻¹ mol⁻¹

The unit of entropy is J K⁻¹ mol⁻¹

Entropy change for the reaction is given as;

ΔS°reaction = ΔS°product - ΔS°reactant

ΔS°reaction = [(4 * 188.8) + (3 * 213.7)] - [269.9 + (5 * 205.1)]

ΔS°reaction = 100.9 J K⁻¹ (mol C₃H₈)⁻¹

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol