Respuesta :
The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ
Further explanation
Enthalpy is the amount of system heat at constant pressure.
The enthalpy is symbolized by H, while the change in enthalpy is the difference between the final enthalpy and the initial enthalpy symbolized by ΔH.
[tex]\large{\boxed{\boxed{\bold{\Delta H=H_{End}-H_{First}}}}[/tex]
Delta H reaction (ΔH) is the amount of heat change between the system and its environment
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The standard unit is kilojoules (kJ)
The enthalpy change symbol (ΔH) is usually written behind the reaction equation.
Change in Standard Evaporation Enthalpy (ΔH vap) is a change in enthalpy at the evaporation of 1 mol liquid phase to the gas phase at its boiling point and standard pressure.
Examples of water evaporation:
H₂O (l) ---> H₂O (g); ΔH vap = + 44kJ
The enthalpy of evaporation is positive because its energy is needed to break the attraction between molecules in a liquid
- 155 g of butane
relative molecular mass of butane (C₄H₁₀) = 4.12 + 10.1 = 58 gram / mol
tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]
[tex]\large mole\:=\:\large \frac{155}{58}[/tex]
mole = 2,672
Since the heat of vaporization for butane is 23.1 kj / mol, the energy needed to evaporate 2,672 moles of butane is:
23.1 kJ / mol x 2,672 mol = 61,723 kJ
Learn more
the heat of vaporization
https://brainly.com/question/11475740
The latent heat of vaporization
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Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic
[tex]\boxed{{\text{61}}{\text{.65 kJ}}}[/tex] of energy will be required to vaporize 155 g of butane.
Further explanation:
Heat of vaporization is the amount of energy that is needed to convert a substance from its liquid state to a gaseous state. It is represented by [tex]\Delta {H_{{\text{vap}}}}[/tex].
The relation between the amount of energy and heat of vaporization is as follows:
[tex]{\text{q}} = {\text{n}}\Delta {H_{{\text{vap}}}}[/tex] …… (1)
Here,
q is the amount of energy.
n is the number of moles of the substance.
[tex]\Delta {H_{{\text{vap}}}}[/tex] is the heat of vaporization.
The moles of butane can be calculated by the following expression:
[tex]{\text{Moles of butane}} = \dfrac{{{\text{Given mass of butane}}}}{{{\text{Molar mass of butane}}}}[/tex] ...... (2)
The given mass of butane is 155 g.
The molar mass of butane is 58.12 g/mol.
Substitute these values in equation (2).
[tex]\begin{aligned}{\text{n}}&= \left( {155{\text{ g}}} \right)\left( {\frac{{1{\text{ mol}}}}{{58.12{\text{ g}}}}} \right)\\&= 2.669{\text{ mol}}\\\end{aligned}[/tex]
The value of n is 2.669 mol.
The value of [tex]\Delta {H_{{\text{vap}}}}[/tex] is 23.1 kJ/mol.
Substitute these values in equation (1) to calculate the energy required for butane.
[tex]\begin{aligned}{\text{q}}&= \left( {{\text{2}}{\text{.669 mol}}} \right)\left( {\frac{{23.1{\text{ kJ}}}}{{1{\text{ mol}}}}} \right)\\&= 61.65{\text{kJ}}\\\end{aligned}[/tex]
Learn more:
- What is the enthalpy of the given reaction? https://brainly.com/question/10412973
- Find the enthalpy of decomposition of 1 mole of MgO: https://brainly.com/question/2416245
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Thermodynamics
Keywords: enthalpy of vaporization, q, n, butane, 58.12 g/mol, 23.1 kJ/mol, 61.65 kJ, 155 g, 2.669 g, amount of energy, moles, molar mass, energy, substance, liquid state, gaseous state.
