Concept explainers
(a)
Interpretation:
Thespecific enthalpy of n-pentane vapor at
Concept introduction:
According to the Hess’s law, the total enthalpy for a reaction is the sum of all the changes in the multiple steps of the reaction. According to this law, enthalpy is a state function.
Specific heat capacity
(1)
Here,
Specific enthalpy
(2)
Here,
(b)
Interpretation:
The specific enthalpy of n-pentane vapor at
Concept introduction:
The specific enthalpy of any substance at its reference temperature is zero. Also, the change in enthalpy between two states of a substance is the difference between the enthalpies of the two states. Thus,
(3)
(c)
Interpretation:
The specific internal energy of the n-pentane vapor at
Concept introduction:
For a
(4)
Here,
The ideal gas law is,
(5)
Here,
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Chapter 8 Solutions
ELEMENTARY PRINCIPLES OF CHEM. PROCESS.
- What is the standard enthalpy of formation of liquid n-butanol, CH3CH2CH2CH20H? CH3CH2CH2CH20H(1) + 602(g) - 4CO2(g) + 5H20(1) AH° = -2675 kJ CO2(g) AH°f (kJ/mol) - 393.5 H20(1) AH°F (kJ/mol) - 285.8 O a. -1996 kJ O b. +3355 kJ O C. -3355 kJ o d. -328 kJ O e. +328 kJarrow_forwardConstruct enthalpy cycles; use Hess's law and the following data to calculate the enthalpy of formation of ethane (from carbon and hydrogen gas). Cis) + Ozig) + CO2(g) AHa = -394kJmol H2g) + %02(a) – H20 m AH°. = -286kJmol1 + 3%O2(g) → 2002(9) + 3H20 m AH = -1560kJmol1arrow_forwardIn a steam boiler, hot gases from a fire transfer heat to water which vapourizes at constant temperature.In certain case, the gases are cooled from 1100°C to 550°C while the water evaporates at 220°C. Thespecific heat of gases is 1.005 kJ/kg K, and the latent heat of water at 220°C is 1858.5 kJ/kg. All the heattransferred from the gases goes to the water. How much does the total entropy of the combined system ofgas and water increase as a result of irreversible heat transfer ? Obtain the result on the basis of 1 kg ofwater evaporated. If the temperature of the surroundings is 30°C find the increase in unavailable energydue to irreversible heat transfer.arrow_forward
- Three moles of nitrogen at 30°C are contained in a container of a certain volume. If we ignore the heat capacity of this container, how much heat should we apply to the system to raise the temperature of nitrogen to 250°C? If the mass of the container is 100 kg and the heat capacity is 0.5 kJ/(kg°C), what is the required heat? At this time, nitrogen is Cv=20.8 and Cp=29.1 J/(mol°C).arrow_forwardc) Calculate the standard enthalpy change for the reaction 3C(s) + 4H2 (g) C;H8 (g) Given that C(s) + O2 (g) CO2 (g) AH° = -394 kJmol·' H2 (g) + ½O2 (g) H2O (I) AH° = -286 kJmol C3H8 (g) + 502 (g) 3CO2 (g) + 4H2O (I) AH° = -2220 kJmol·'arrow_forwardThe standard enthalpy change of combustion [to CO2(g) and H2O()] at 25°C of the organic solid isophthalic acid, C8H6O4(s), is determined to be -3185.8 kJ mol-1. What is the Hf° of C8H6O4(s) based on this value?arrow_forward
- Solve using gas constant (R) of the mixture. Thank youarrow_forwarda) What is the standard enthalpy of reaction per mole of either HCI or NaOH when 100.00 cm³ of 0.1 moldm-³ of hydrochloric acid (HCI) and 100.00 cm³ of 0.1 moldm-3 of sodium hydroxide (NaOH) are mixed? According to your simulation, the temperature rose by 0.68 Specific heat capacity of water = 4.18 Jg-¹ °C-1) Show your working below: HCI (ag) + NaOH(aq) → NaCl(aq) + H₂Om °C. (Density of water = 1gcm-³, Iarrow_forwardA coffee cup calorimeter with a heat capacity of 7.40 JrC was used to measure the change in enthalpy of a precipitation reaction. A 50.0 mL solution of 0.360 M AGNO, was mixed with 50.0 mL of 0.310 M KBr. After mixing, the temperature was observed to increase by 3.04 °C. Calculate the enthalpy of reaction, AHan, per mole of precipitate formed (AgBr). Assume the specific heat of the product solution is 4.11 J/ (g C) and that the density of both the reactant solutions is 1.00 g/mL. Calculate the theoretical moles of precipitate formed from AgNO, and KBr. theoretical moles of precipitate formed from AgNO,: moles theoretical moles of precipitate formed from KBr: moles Calculate the heat change experienced by the calorimeter contents, qontents -arrow_forward
- What is AG (in kJ) for 2 SO2(g) + O2(g) – 2 S03(g) at 700 K, under standard conditions of 1 bar partial pressure for all gases, given that K = 3.0×10“ at 700 K? O -60. +36 0.0 +60. -36arrow_forwardPlease don't provide handwritten solution .....arrow_forward4.00 g NaOH(s) (39.996 g mol–1) dissolves in 100.0 g of water in a constant–pressure calorimeter. The initial temperatures of both materials are the same at 20.0°C. The density of the solution is 1.00 g cm–3. The heat of dissolution is –44.51 kJ mol–1. The specific heat of the solution is 4.184 J g–1 °C–1. The specific heat of the calorimeter is 100 J °C–1. What is the final temperature of the solution? 30.23°C 25.94°C 28.32°C What is the energy lost to the calorimeter? 579 J 832 J 721 Jarrow_forward
- Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning