Verified Answer :
m=2.65kg=2650g
M=74g/mol
n = m/M = 2650g/74g/mol ≈ 35.81mol
Heat of fusion = 7.27kJ/mol
If 7.27kJ ------------- energy is in --------------- 1mol
than x ------------- energy is in --------------- 35.81mol
x = [35.81mol*7.27kJ]/1mol = 260.3387kJ
The amount of energy released when \({{\text{(}}{{\text{C}}_2}{{\text{H}}_5}{\text{)}}_2}{\text{O}}\) freezes is \(\boxed{{\text{259}}{\text{.92 kJ}}}\).
Further explanation:
Heat of fusion measures the amount of energy needed to heat given mass of a solid to change its state into liquid. Also, it represents the amount of energy given up when a given mass of liquid solidifies.
When a substance converts from its phase from solid to liquid, the change in enthalpy \(\left({\Delta H}}}\right)\) is positive, and if the substance transforms from liquid state to solid state then the change in enthalpy\(\left({\Delta H}}}\right)\) is negative.
The expression to relate the energy released and heat of solidification is as follows:
\({\text{Energy released}}={\text{n}}\times\left({{\Delta }}{{\text{H}}_{{\text{solidification}}}}}\right)\)
Given mass of \({{\text{(}}{{\text{C}}_2}{{\text{H}}_5}{\text{)}}_2}{\text{O}}\) is 2.65 kg.
The conversion factor to convert mass in (g) from (kg) is written as follows:
\({\text{1kg}}=1000\;{\text{g}}\)
So, 2.65 kg of mass is converted into (g) as follows:
\(\begin{aligned}{\text{Mass}}&=\,\left({2.65\;{\text{kg}}} \right)\left( {\frac{{{\text{1000}}\;{\text{g}}}}{{{\text{1}}\;{\text{k}}{\text{g}}}}}\right)\\&={\text{2650 g}}\\\end{aligned}\)
Molar mass of \({{\text{(}}{{\text{C}}_2}{{\text{H}}_5}{\text{)}}_2}{\text{O}}\) is 74.12 g/mol.
The moles of \({{\text{(}}{{\text{C}}_2}{{\text{H}}_5}{\text{)}}_2}{\text{O}}\)can be calculated as follows:
\(\begin{aligned}{\text{number of moles }}&=\frac{{{\text{mass of (}}{{\text{C}}_2}{{\text{H}}_5}{{\text{)}}_2}{\text{O}}}}{{{\text{molar mass of (}}{{\text{C}}_2}{{\text{H}}_5}{{\text{)}}_2}{\text{O}}}}\\&=\frac{{{\text{2650 g}}}}{{{\text{74}}{\text{.12}}\;{\text{g/mol}}}}\\&=35.752{\text{ mol}}\\\end{aligned}\)
The heat of fusion for diethyl ether\({{\text{(}}{{\text{C}}_2}{{\text{H}}_5}{\text{)}}_2}{\text{O}}\)is 7.27 kJ/mol and therefore the heat of solidification for diethyl ether\({{\text{(}}{{\text{C}}_2}{{\text{H}}_5}{\text{)}}_2}{\text{O}}\)is -7.27 kJ/mol.
The formula to calculate the amount of energy released is as follows:
\({\text{Amount of energy}}={\text{number of moles}}\times{\Delta }}{{\text{H}}_{{\text{solidification}}}}{\text{ }}\) …… (1)
Substitute 35.752 mol for number of moles and -7.27 kJ/mol for \({\Delta }}{{\text{H}}_{{\text{solidification}}}}\) in the equation (1).
\(\begin{aligned}{\text{Amount of energy}}&=\left({35.752\;{\text{mol}}}\right)\left({-7.27{\text{kJ/mol}}}\right)\\&=-259.92{\text{ kJ}}\\\end{aligned}\)
Therefore, the amount of energy released is \({\mathbf{259}}{\mathbf{.92 kJ}}\).
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Thermodynamics
Keywords: Heat of fusion, number of moles, enthalpy, enthalpy change, released, absorbed, 259.91 kJ, 7.27kJ/mol, and 35.752 mol.