5/7/2023 0 Comments Workdone abIndicates the work is done on the system.Ģ x (12 + 16) = 56 g of CO on oxidation liberates 566 kJ energy How much pressure volume work is done and what is the value of ΔU for the reaction of 7.0 g of CO at 1 atm pressure, if the volume change is – 2.8 L.ĢCO (g) + O 2(g) → 2CO 2(g) Enthalpy change = Δ H = – 566 kJ Indicates that work is done by the system on the surroundingsĪns: Work done by the surroundings on the system in the reaction is – 18.61 kJ.ĬO reacts with O2 according to the following reaction. State whether work is on the system or by the system. W = – Δn RT = – (-½) mol × 8.314 J K -1 mol -1 × 323 K =Īns: Work done by the surroundings on the system in the reaction is 1343 JĬalculate the work done in the following reaction when 2 mol of NH 4NO 3 decomposes at constant pressure at 10o ☌. Is used, hence dividing equation by 2 to get 1 mol of SO 2 The work done in the following reaction when 1 mol of SO 2 is Indicates that work is done by the surroundings on the systemĪns: Work done by the surroundings on the system in the reaction is 1758 J HCl are used, hence dividing equation by 2 to get 2 HCl, we get State whether work is on the system or by the The work done in the following reaction when 2 moles of HCl are used atĬonstant pressure and 423 K. Process is given by W = – P ext × ΔV = – P ext ×Ĥ000 kJ (Heat absorbed by sample), W = + 2000 kJ (Work done by surroundings)Ĥ000 kJ (Heat absorbed by sample), W = – 600 kJ (Work done on the surroundings) What is ΔU?Ĥ000 kJ (Heat absorbed by sample), ΔV = 0 (Volume remains the same) The original sample absorbs heat, it expands against atmospheric pressure andĭoes 600 kJ of work on its surroundings. Surrounding does 2000 kJ of work on the sample. ΔU? b) Suppose that in addition to absorption of heat by the sample, the = Heat supplied at constant pressure = 2444 kJ Ideal gas expands from a volume of 6 dm³ to 16 dm³ against constant external = Heat supplied at constant pressure = + 6 kJĪns: The change in internal energy is 4.5 kJ and enthalpy change is 6 kJ In internal energy = ΔU =? Enthalpy change = ΔH =? What are the change in internal energy and enthalpy changeĦ kJ (Heat absorbed), W = -1.5 kJ (Work done on the surroundings). A particular reaction, the system absorbs 6 kJ of heat and does 1.5 kJ of work
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