![]() Calculate the heat released upon formation of 35.2 g ofCO 2 from carbon and dioxygen gas.įormation of CO 2 from carbon and dioxygen gas can be represented as: Enthalpy of combustion of carbon to CO 2 is –393.5−7.151kJmol −1 . Hence, the enthalpy change involved in the transformation is−7.151kJmol −1.ġ1. (c) Energy change involved in the transformation of 1 mol of ice at0∘C to 1 mol of ice at−10 ∘C. ![]() (b) Energy change involved in the transformation of 1 mol of water at 0∘ to 1 mol of ice at 0 ∘C (a) Energy change involved in the transformation of 1 mol of water at 10 ∘ C to 1 mol of water at 0 C. Total enthalpy change involved in the transformation is the of the following changes: Calculate the enthalpy change on freezing of 1.0 mol of water at10.0☌ to ice at Substituting the values in the expression of q:ġ0. Molar heat capacity of Al is 24 Jmol −1K −1.įrom the expression of heat (q) q=m⋅ c. ΔT Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35☌ to 55☌. Substituting the values in the expression of ΔH: NH 2CN(g) + 3/2O 2(g)→N 2(g)+CO 2(g)+H 2O(l)Įnthalpy change for a reaction (ΔH) is given by the expression, ![]() Calculate enthalpy change for the reaction at 298 K. The reaction of cyanamide,NH 2CN(s) with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7kJmol −1 at 298 K. Hence, the change in internal energy for the given process is 307 J.Ĩ. Substituting the values in expression (i), we get W = –394 J (Since work is done by the system) ΔU = change in internal energy for a process What is the change in internal energy for the process?Īccording to the first law of thermodynamics, In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. Therefore, the reaction is spontaneous at any temperature.ħ. The reaction will beįor a reaction to be spontaneous, ΔG should be negative.Īccording to the question, for the given reaction, A reaction, A + B → C + D + q is found to have a positive entropy change. ∴ Enthalpy of formation of CH 4(g)=−74.8kJmol −1 Hence, alternative (i) is correct.Ħ. Thus, the desired equation is the one that represents the formation of CH 4(g) i.e., The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3kJmol −1−393.5kJmol −1, and −285.8kJmol −1 The enthalpies of all elements in their standard states are:ĥ. Hence, under adiabatic conditions, q = 0.ģ. Solution : A system is said to be under adiabatic conditions if there is no exchange Of heat between the system and its surroundings. For the process to occur under adiabatic conditions, the correct condition is: depend only on the state of a system and not on the path.Ģ. (iv) whose value depends on temperature onlyĪ thermodynamic state function is a quantity Whose value is independent of a path. (iii) used to determine pressure volume work A thermodynamic state function is a quantity NCERT SOLUTIONS FOR CLASS 11 CHEMISTRY CHAPTER 6- THERMODYNAMICS – Exercises Chapter-6 Thermodynamics Answer the following Questions.ġ. Important exam-based questions are covered in depth in each chapter. Use the Swastik Classes’ NCERT answers for Chemistry class 11 as a resource. One should not omit any NCERT textbook content in order to get the highest possible grade. The NCERT answers for class 11 Chemistry were created with the goal of providing students with the most help possible.Īnswers to the class 11 Chemistry questions provided in the exercise might be challenging for students for a number of reasons. ![]() In general, Class 11 is regarded as the most significant year in a student’s professional development. Top students love SWC NCERT Solutions because they are very effective. The NCERT Solution for Class 11 Chemistry includes answers to every question from the NCERT text book’s exercise.
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