Answer:
[tex]\large \boxed{\Delta E_{\text{rxn}}= \text{-5160 kJ$\cdot$ mol}^{-1}}}[/tex]
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
[tex]n = \text{1.025 g} \times \dfrac{\text{1 mol}}{\text{128.17 g}} = 7.997 \times 10^{-3}\text{ mol}[/tex]
(b) ΔT
ΔT = T_f - Ti = 32.33 °C - 24.25 °C = 8.08 °C
(c) ΔE
[tex]\begin{array}{ccccl}n\Delta E & +& C_{cal} \Delta T&=&0\\7.997 \times 10^{-3}\text{ mol}\times \Delta E& + & 5.11 \text{ kJ$^{\circ}$C$^{-1}$} \times 8.08 \, ^{\circ}\text{C} & = & 0\\7.997 \times 10^{-3}\Delta E \text{ mol} & + & \text{41.29 kJ} & = & 0\\&&7.997 \times 10^{-3}\Delta E \text{ mol} & = & -\text{41.29 kJ} & & \\\end{array}\\[/tex]
[tex]\begin{array}{ccccl}& &\Delta E & = & -\dfrac{\text{41.29 kJ}}{7.997 \times 10^{-3}\text{ mol}}\\\\& &\Delta E & = & \mathbf{-5160} \textbf{ kJ}\cdot\textbf{mol}^{\mathbf{-1}}\\\end{array}\\\text{The change in energy is $\large \boxed{\textbf{-5160 kJ$\cdot$ mol}^{\mathbf{-1}}}$}[/tex]
Answer:
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
