Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.025 gof naphthalene is burned in a bomb calorimeter, the temperature rises from 24.25 ∘C to 32.33 ∘C. Find ΔErxn for the combustion of naphthalene. The heat capacity of the calorimeter, determined in a separate experiment, is 5.11kJ/∘C. Express the change in energy in kilojoules per mole to three significant figures.ΔErxn = kJ/mol

Explanation:

There are two energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm calorimeter .

Per the Law of Conservation of Energy,

q₁ + q₂ = 0

Data:

For the reaction

mass of naphthalene = 1.025 g

For the calorimeter

Ccal = 5.11 kJ·°C⁻¹; Ti = 24.25 °C; T_f = 32.33 °C

Calculations

(a) Moles of naphthalene

(b) ΔT

ΔT = T_f - Ti = 32.33 °C - 24.25 °C = 8.08 °C

(c) ΔE

ΔErxn = -5161 kJ/mol

Explanation:

Step 1: Data given

Mass of naphthalene = 1.025 grams

The temperature rises from 24.25 °C to 32.33 °C

Molar mass of napthalene = 128.17 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles naphthalene = 1.025 grams / 128.17 g/mol

Moles naphthalene = 0.0080 moles

Step 3: Calculate the amount of heat released

heat = (5.11 kJ/°C)(32.33 °C - 24.25 °C) = 41.3 kJ

Since there is heat released, this is negative = -41.3 kJ

Step 4: Calculate  Δ(Hrxn) per mole of naphthalene is:

H = (-41.3 kJ)/(0.0080 moles) = -5161 kJ/mol

ΔErxn = -5161 kJ/mol

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