{"id":40074,"date":"2024-02-19T08:03:04","date_gmt":"2024-02-19T02:33:04","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=40074"},"modified":"2024-02-19T12:45:21","modified_gmt":"2024-02-19T07:15:21","slug":"plus-one-chemistry-chapter-wise-previous-questions-chapter-6","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/plus-one-chemistry-chapter-wise-previous-questions-chapter-6\/","title":{"rendered":"Plus One Chemistry Chapter Wise Previous Questions Chapter 6 Thermodynamics"},"content":{"rendered":"

Plus One Chemistry Chapter Wise\u00a0Previous Questions\u00a0Chapter 6 Thermodynamics <\/strong>is part of Kerala Plus One Chemistry Chapter Wise Previous Year Questions and Answer<\/a> . Here we have given Plus One Chemistry Previous Questions Chapter 6 Thermodynamics.<\/p>\n

Kerala Plus One Chemistry Chapter Wise Previous Questions Chapter 6 Thermodynamics<\/h2>\n

Question 1.
\na) State Hess\u2019s Law of constant heat summation. (March – 2009)<\/span>
\nb) The equilibrium constant for a reaction is 5. What will be the value of \u0394G\u00b0.
\nGiven that R = 8.314JK-1<\/sup> mol-1<\/sup>, T = 300K.
\nR = 8.314 JK-1<\/sup> mol-1<\/sup>
\nT = 300K
\nAnswer:
\na) Hess\u2019s law : Enthalpy change in a chemical reaction is same whether it takes place in one step or more than one step,
\nb) \u0394G = -2.303RT logK
\n\u0394G = -2.303 x 8.314 x 300 x log5
\n= -2.30 x 8.314 x 300 x 0.6989
\n= -4014.581 J\/mol<\/p>\n

Question 2.
\nA system in thermodynamics refers to that part of the universe in which observation are made. (March – 2010)<\/span>
\na) What do you mean by an isolated system? Given an example.
\nb) Distinguish between intensive and extensive properties. Give two examples for each.
\nAnswer:
\na) A system which can neither exchange matter nor energy with the surrounding. It is called as isolated system.
\nEg: Thermo flask
\nb) Extensive property : The properties of the system which depend upon the quantity of the matter contained in the system.
\nEg: mass, volume.
\nIntensive property : The properties of the system whose values are independent of the quantity of substance present in the system.
\nEg: Temperature, pressure.<\/p>\n

Question 3.
\nLattice enthalpy of an ionic salt is a factor that determine its stability. (Say – 2010)<\/span>
\na) Define the lattice enthalpy.
\nb) Draw the Born-Haber cycle for the calculation of lattice enthalpy of the ionic crystal NaCI.
\nAnswer:
\na) The lattice enthalpy of an ionic solid is defined as the energy required to completely separate one mole of the ionic compound into gaseous constituent ions.
\nb) Na(s)<\/sub> + 1\/2CI2(g)<\/sub> \u2192 NaCI(s)
\n<\/sub>\"Plus<\/p>\n

Question 4.
\nThe spontaneity of a process is expressed in terms of a change in Gibbs energy. (March – 2011)<\/span>
\na) What is meant by a change in Gibbs energy of a system?
\nb) How is it related to the enthalpy and entropy of a system?
\nc) How is it useful in predicting the feasibility of a process?
\nAnswer:
\na) Free energy is the energy available to a system that can be converted into useful work.
\nb) \u0394G = \u0394H – T\u0394S
\nc) For a spontaneous process, \u0394G = –<\/sup>ve
\nFor Non-Spontaneous process \u0394G = +<\/sup>ve
\nIf the system is at equilabrium \u0394G = O<\/p>\n

Question 5.
\nA spontaneous process is an irreversible process and may only by reserved by some external agency. (Say – 2012)<\/span>
\na) Decrease in enthalpy is the only criterion for spontaneity. Do you agree? Why?
\nb) Calculate the work done for the reversible iso thermal expansion of 1 mole of an ideal gas at 27\u00b0C, from a volume of 10 dm3 to a volume of 20dm3<\/sup>.
\nAnswer:
\na) No. Decrease in enthalpy (\u0394H = \u2014ve) is not the only criterion for spontaneity. There are many endothermic processes which are also spontaneous like the evaporation of water placed in an open vessel. A process takes place spontaneously not only in the direction in which enthalpy of the system is reduced but also in the direction in which disorder or randomness increases (\u0394S = +ve ). Thus, a resultant of the tendencies to decrease enthalpy and increase entropy determines the spontaneity of a process.<\/p>\n

\"Plus<\/p>\n

Question 6.
\nThermodynamics deals with energy changes of macroscopic systems. (March – 2012)<\/span>
\na) Consider a chemical reaction taking place in a dosed insulated vessel. To which type of ther modynamic system does it belong?
\nb) State the first law of thermodynamics.
\nc) 3 moI of an ideal gas at 1.5 atm and 25\u00b0C expands isothermally in a reversible mannerto twice its original volume against an external pressure of 1 atm. Calculate the work done. [R = 8.314 JK-1<\/sup> mol-1<\/sup>]
\nAnswer:
\na) Isolated system
\nb) Ist<\/sup> law of thermodynamics states that \u2018the energy of an isolated system is Constant\u2019.
\n\"Plus<\/p>\n

Question 7.
\na) i) Construct an enthalpy diagram for the determination of lattice enthalpy of sodium chloride. (Say – 2012)<\/span>
\nii) Enthalpy and entropy changes of a reaction are40.63kJ\/mol and 108.8 Jk-1<\/sup> mol-1<\/sup>. Predict the feasibility of the reaction atr 27\u00b0C.
\nOR
\nb) i) Explain the Hess\u2019s law of constant heat summation, with an example.
\nii) Draw the enthalpy diagram for an exothermic and endothermic reaction.
\nAnswer:
\n\"Plus
\nSince \u0394G is +ve, the reaction is not feasible at 27\u00b0C.
\nOR
\nb) i) Hes&s law states that the enthalpy change during a process is the same whether \u00a1t takes place in one step or in several steps.<\/p>\n

Then according to Hess\u2019s law
\n\"Plus<\/p>\n

Question 8.
\nMost of the naturally occurring processes are spontaneous. (March – 2013)<\/span>
\na) Give the criteria for spontaneity of a process in terms of free energy change (\u0394G)
\nb) Exothermic reactions associated with a decrease in entropy are spontaneous at lower temperatures. Justify on the basis of Gibbs equation.
\nc) Find the temperature above which the reaction
\n\"Plus
\nAnswer:
\na) Free energy change (\u0394G) is negative for a spontaneous process.
\nb) Far a Process to be spontaneous \u0394G should be -ve. But \u0394G = \u0394H -T \u0394S ………….. (1)
\nFor Exothermic reaction \u0394H =’ve
\nand decrease in entropy \u0394S = -ve.
\nSubstitute the value of \u0394H and \u0394S in (1) \u0394G become -ve only when \u0394H > T \u0394S. Such reactions are spontaneous only at low temperatures.
\nAt 2474.7K and above this temperature the reaction become spontaneous.<\/p>\n

\"Plus<\/p>\n

Question 9.
\na) The enthalpy of combustion of CH4(g)<\/sub>, C(Graphite)<\/sub> and H2(g)<\/sub> at 298 K are -890.3 kJ mor-1<\/sup>, -393.5 kJ mor-1<\/sup> and -285.8 kJ moM respectively. <\/span>Calculate the enthalpy of formation of CH4(g)<\/sub>. (Say – 2013)<\/span>
\nb) Match the following:
\n\"Plus
\nAnswer:
\n\"Plus
\n\"Plus<\/p>\n

Question 10.
\na) For the oxidation of iron 4 Fe(s)<\/sub> + 3O2(g)<\/sub> \u2192 2 Fe2<\/sub>O3(s)<\/sub>, entropy change is -549.4 JK-1\u00a0<\/sup>mol-1<\/sup> at 298 K. Inspite of the negative entropy change of this reaction, why is the reaction spontaneous? (\u0394H\u00b0r<\/sub>\u00a0for the reaction is -1648 x 10 Jmol-1<\/sup>). (March – 2014)<\/span>
\nb) Write the dfference between extensive and intensive properties. Give one example of each.
\nAnswer:
\n\"Plus
\nb) Extensive properties – properties whose values depend on the quantity or size of matter present in the system.
\ne.g., Mass, Volume. Internal energy. Enthalpy, Heat capacity etc.<\/p>\n

Intensive properties- properties which do not depend on the quantity or size of matter present in the system.
\ne.g., Temperature, Density, Pressure etc.<\/p>\n

Question 11.
\na) \u0394G gives a criteria for spontaneity of reactions at a constant pressure and temperature. How is \u0394G helpful in predicting the spontaneity of the reaction? (August – 2014)<\/span>
\nb) State and explain Hess\u2019s law of constant heat summation.
\nAnswer:
\na) If \u0394G is negative (<O), the proces is spontaneous. If \u0394G is posit\u00eeve (>0), the proces is non-spontaneous.
\nb) Hess\u2019s law states that the enthalpy change during a process is the same whether it takes place in one step or in several steps.<\/p>\n

Then according to Hess\u2019s law
\n\"Plus<\/p>\n

Question 12.
\na) Classify the following into intensive and extensive properties. (March – 2015)<\/span>
\ni) Inernal energy
\nii) Density
\nii) Heat Capacity
\niv) Temperature
\nb) Calculate the standard free energy change (\u0394G\u00b0) for the conversion of oxygen to ozone 3\/2 O2(g)<\/sub> \u2192 O3(g)<\/sub> at 298 K if the equilibrium constant for the conversion is 2.47 x 10-29<\/sup>. (Given R = 8.314 JK-1<\/sup> mol-1<\/sup>)
\nAnswer:
\na) Intensive properties –
\nii) Density and
\niv) Temperature<\/p>\n

Extensive properties-
\ni) Internal energy
\niii) Heat capacity<\/p>\n

\"Plus<\/p>\n

Question 13.
\nExpansion of a gas in vacuum is called free expansion. (Say – 2015)<\/span>
\na) Which one of the following represents free expansion of an ideal gas under adiabatic conditions?
\ni) q = 0, \u0394T \u2260 0, w = 0
\nii) q \u2260 0, \u0394T = 0, w= 0
\niii) q = 0, \u0394T = 0, w = 0
\niv) q = 0, \u0394T < 0, w = 0
\nb) The enthalpy change for the reaction.
\n\"Plus
\nAnswer:
\n\"Plus<\/p>\n

Question 14.
\nThe enthalpy change in a process is the same, whether the process is carried out in a single step or inseveral steps. (March – 2016)<\/span>
\na) Identify the law stated here.
\nb) Calculate the enthalpy of formation of CH4<\/sub> from the following data:
\n\"Plus
\nAnswer:
\na) Hess\u2019s law
\nb) The required equation is
\n\"Plus
\nThe given data are:
\n\"Plus<\/p>\n

Question 15.
\na) Which of the following is a process taking place with increase in entropy? (Say – 2016)<\/span>
\ni) Freezing of water
\nii) Condensation of steam
\niii) Cooling of a liquid
\niv) Dissolution of a solute
\nb) State and illustrate Hess\u2019s law.
\nAnswer:
\na) iv) Dissolution of a solute
\nb) Hess\u2019s law states that if a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature. Illustration:<\/p>\n

Consider the formation of CO2<\/sub> from carbon and oxygen. There are two ways by which the change can be brought about.
\n\"Plus
\nIn general, if enthalpy of an overall reaction A \u2192 B along one route is \u03941<\/sub>H\u00a0and \u03941<\/sub>H1<\/sub>, \u03942<\/sub>H2<\/sub>, \u03943<\/sub>H3\u00a0<\/sub>…. representing enthalpies of reactions leading to same product, B along another route, then according to Hess\u2019s law,
\n\"Plus<\/p>\n

Question 16.
\na) Some macroscopic properties are given below. (March – 2017)<\/span>
\nHelp Reena to classify then into two groups under suitable titles [Heat capacity, Entropy, Refractive index, Surface tension] (2)
\nb) For the reaction
\n2A(g)<\/sub> + B(g)<\/sub> \u2192 2D(g)<\/sub>
\nAU\u00b0 = -10.5kJ\/mol
\n\u0394S\u00b0 = -44.1J\/k\/mol at 298k.
\nCalculate AG\u00b0forthe reaction
\nAnswer:
\na) Extensive properties: Heat capacity, Entropy Intensive properties: Refractive index, Surface tension
\n\"Plus<\/p>\n

We hope the Kerala Plus One Chemistry Chapter Wise Previous Questions Chapter 6 Thermodynamics help you. If you have any query regarding Kerala Plus One Chemistry Chapter Wise Previous Questions Chapter 6 Thermodynamics, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

Plus One Chemistry Chapter Wise\u00a0Previous Questions\u00a0Chapter 6 Thermodynamics is part of Kerala Plus One Chemistry Chapter Wise Previous Year Questions and Answer . Here we have given Plus One Chemistry Previous Questions Chapter 6 Thermodynamics. Kerala Plus One Chemistry Chapter Wise Previous Questions Chapter 6 Thermodynamics Question 1. a) State Hess\u2019s Law of constant heat […]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[42728],"tags":[],"yoast_head":"\nPlus One Chemistry Chapter Wise Previous Questions Chapter 6 Thermodynamics - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/plus-one-chemistry-chapter-wise-previous-questions-chapter-6\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Plus One Chemistry Chapter Wise Previous Questions Chapter 6 Thermodynamics\" \/>\n<meta property=\"og:description\" content=\"Plus One Chemistry Chapter Wise\u00a0Previous Questions\u00a0Chapter 6 Thermodynamics is part of Kerala Plus One Chemistry Chapter Wise Previous Year Questions and Answer . 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Here we have given Plus One Chemistry Previous Questions Chapter 6 Thermodynamics. Kerala Plus One Chemistry Chapter Wise Previous Questions Chapter 6 Thermodynamics Question 1. a) State Hess\u2019s Law of constant heat […]","og_url":"https:\/\/www.aplustopper.com\/plus-one-chemistry-chapter-wise-previous-questions-chapter-6\/","og_site_name":"A Plus Topper","article_publisher":"https:\/\/www.facebook.com\/aplustopper\/","article_published_time":"2024-02-19T02:33:04+00:00","article_modified_time":"2024-02-19T07:15:21+00:00","og_image":[{"url":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Plus-One-Chemistry-Chapter-Wise-Previous-Questions-Chapter-6-Thermodynamics-1.png"}],"twitter_card":"summary","twitter_misc":{"Written by":"Kalyan","Est. reading time":"8 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/www.aplustopper.com\/#organization","name":"Aplus Topper","url":"https:\/\/www.aplustopper.com\/","sameAs":["https:\/\/www.facebook.com\/aplustopper\/"],"logo":{"@type":"ImageObject","@id":"https:\/\/www.aplustopper.com\/#logo","inLanguage":"en-US","url":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg","contentUrl":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg","width":1585,"height":375,"caption":"Aplus Topper"},"image":{"@id":"https:\/\/www.aplustopper.com\/#logo"}},{"@type":"WebSite","@id":"https:\/\/www.aplustopper.com\/#website","url":"https:\/\/www.aplustopper.com\/","name":"A Plus Topper","description":"Improve your Grades","publisher":{"@id":"https:\/\/www.aplustopper.com\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/www.aplustopper.com\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/www.aplustopper.com\/plus-one-chemistry-chapter-wise-previous-questions-chapter-6\/#primaryimage","inLanguage":"en-US","url":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Plus-One-Chemistry-Chapter-Wise-Previous-Questions-Chapter-6-Thermodynamics-1.png","contentUrl":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Plus-One-Chemistry-Chapter-Wise-Previous-Questions-Chapter-6-Thermodynamics-1.png","width":279,"height":240,"caption":"Plus One Chemistry Chapter Wise Previous Questions Chapter 6 Thermodynamics 1"},{"@type":"WebPage","@id":"https:\/\/www.aplustopper.com\/plus-one-chemistry-chapter-wise-previous-questions-chapter-6\/#webpage","url":"https:\/\/www.aplustopper.com\/plus-one-chemistry-chapter-wise-previous-questions-chapter-6\/","name":"Plus One Chemistry Chapter Wise Previous Questions Chapter 6 Thermodynamics - 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