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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2 answers:
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 .
The relation between the amount of energy and heat of vaporization is as follows:
…… (1)
Here,
q is the amount of energy.
n is the number of moles of the substance.
is the heat of vaporization.
The moles of butane can be calculated by the following expression:
...... (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).
The value of n is 2.669 mol.
The value of is 23.1 kJ/mol.
Substitute these values in equation (1) to calculate the energy required for butane.
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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.
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.
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]
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
<h3>Learn more</h3>the heat of vaporization
The latent heat of vaporization
Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic
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Answer:
The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.
Explanation:
The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.
For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:
In this case:
Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:
=2.09*10⁻²
=4.14*10⁻²
= 0.280
So,
Qc= 0.011
Comparing Qc with Kc allows to find out the status and evolution of the system:
If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.
Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. <u><em>The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.</em></u>
The moles of gas in the bottle has been 0.021 mol.
The ideal gas has been given as the gas where there has been negligible amount of interatomic collisions. The ideal gas equation has been given as:
The gi<em>ve</em>n gas has standard pressure,
The volume of the gas has been,
The temperature of the gas has been,
Substituting the values for the moles of gas, <em>n:</em>
<em />
<em />
The moles of gas in the bottle has been 0.021 mol.
Learn more about ideal gas, here:
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