How much energy is required to vaporize 155 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol

Question

Gwar [14]

4 years ago

6

How much energy is required to vaporize 155 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol

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Chemistry

2 answers:

klemol [59] · Answer 1 · 2021-11-09T13:38:52+03:00

$\boxed{{\text{61}}{\text{.65 kJ}}}$ 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 $\Delta {H_{{\text{vap}}}}$ .

The relation between the amount of energy and heat of vaporization is as follows:

${\text{q}} = {\text{n}}\Delta {H_{{\text{vap}}}}$ …… (1)

Here,

q is the amount of energy.

n is the number of moles of the substance.

$\Delta {H_{{\text{vap}}}}$ is the heat of vaporization.

The moles of butane can be calculated by the following expression:

${\text{Moles of butane}} = \dfrac{{{\text{Given mass of butane}}}}{{{\text{Molar mass of butane}}}}$ ...... (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).

$\begin{aligned}{\text{n}}&= \left( {155{\text{ g}}} \right)\left( {\frac{{1{\text{ mol}}}}{{58.12{\text{ g}}}}} \right)\\&= 2.669{\text{ mol}}\\\end{aligned}$

The value of n is 2.669 mol.

The value of $\Delta {H_{{\text{vap}}}}$ is 23.1 kJ/mol.

Substitute these values in equation (1) to calculate the energy required for butane.

$\begin{aligned}{\text{q}}&= \left( {{\text{2}}{\text{.669 mol}}} \right)\left( {\frac{{23.1{\text{ kJ}}}}{{1{\text{ mol}}}}} \right)\\&= 61.65{\text{kJ}}\\\end{aligned}$

Learn more:

What is the enthalpy of the given reaction? brainly.com/question/10412973
Find the enthalpy of decomposition of 1 mole of MgO: 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.

netineya [11] · Answer 2 · 2021-11-09T12:24:39+03:00

The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ

<h3>Further explanation</h3>

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.

$\large{\boxed{\boxed{\bold{\Delta H=H_{End}-H_{First}}}}$

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